strToHex ( string to it's hex representation as string)Integer-to-hex string generatorConvert hex string to...
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strToHex ( string to it's hex representation as string)
Integer-to-hex string generatorConvert hex string to byte arrayString representation of TreeSimple string inverter programSimple string repeater programGeneric Pairing Heap PerformanceHex-string representation of a byte arrayConvert hex color string to SDL ColorC binary number converter (Hex and Decimals)Convert string of hex into vector of bytes
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}
$begingroup$
I want to convert strings to it's hex representation as string too (like hex dump programs), for example "abz"
to "61627A"
char * strToHex( char * str )
{
int length = strlen ( str );
char * newStr = malloc( length * 2 );
if ( !newStr ) shutDown ( "can't alloc memory" ) ;
for ( int x = 0; x < length; x++){
char y = str[ x ];
sprintf ( newStr + x * 2, "%02X", y );
}
return newStr;
}
ShutDown
definition is omitted here, it is a function that calls perror
and exit()
I designed strToHex
to be used like
char * str = "abcdefghijklmnopqrstuvwxyz";
char * hex = strToHex(str);
printf("%sn",hex);
//outputs : 6162636465666768696A6B6C6D6E6F707172737475767778797A
beginner c strings
$endgroup$
add a comment |
$begingroup$
I want to convert strings to it's hex representation as string too (like hex dump programs), for example "abz"
to "61627A"
char * strToHex( char * str )
{
int length = strlen ( str );
char * newStr = malloc( length * 2 );
if ( !newStr ) shutDown ( "can't alloc memory" ) ;
for ( int x = 0; x < length; x++){
char y = str[ x ];
sprintf ( newStr + x * 2, "%02X", y );
}
return newStr;
}
ShutDown
definition is omitted here, it is a function that calls perror
and exit()
I designed strToHex
to be used like
char * str = "abcdefghijklmnopqrstuvwxyz";
char * hex = strToHex(str);
printf("%sn",hex);
//outputs : 6162636465666768696A6B6C6D6E6F707172737475767778797A
beginner c strings
$endgroup$
2
$begingroup$
I'd be really interested to see what shutdown(char* msg) does.
$endgroup$
– pacmaninbw
1 hour ago
$begingroup$
In the use case that was provided, since you can effectively predict the size, I would think it would be more natural to have a string buffer and the size passed in instead of creating it dynamically.
$endgroup$
– Neil Edelman
28 mins ago
add a comment |
$begingroup$
I want to convert strings to it's hex representation as string too (like hex dump programs), for example "abz"
to "61627A"
char * strToHex( char * str )
{
int length = strlen ( str );
char * newStr = malloc( length * 2 );
if ( !newStr ) shutDown ( "can't alloc memory" ) ;
for ( int x = 0; x < length; x++){
char y = str[ x ];
sprintf ( newStr + x * 2, "%02X", y );
}
return newStr;
}
ShutDown
definition is omitted here, it is a function that calls perror
and exit()
I designed strToHex
to be used like
char * str = "abcdefghijklmnopqrstuvwxyz";
char * hex = strToHex(str);
printf("%sn",hex);
//outputs : 6162636465666768696A6B6C6D6E6F707172737475767778797A
beginner c strings
$endgroup$
I want to convert strings to it's hex representation as string too (like hex dump programs), for example "abz"
to "61627A"
char * strToHex( char * str )
{
int length = strlen ( str );
char * newStr = malloc( length * 2 );
if ( !newStr ) shutDown ( "can't alloc memory" ) ;
for ( int x = 0; x < length; x++){
char y = str[ x ];
sprintf ( newStr + x * 2, "%02X", y );
}
return newStr;
}
ShutDown
definition is omitted here, it is a function that calls perror
and exit()
I designed strToHex
to be used like
char * str = "abcdefghijklmnopqrstuvwxyz";
char * hex = strToHex(str);
printf("%sn",hex);
//outputs : 6162636465666768696A6B6C6D6E6F707172737475767778797A
beginner c strings
beginner c strings
edited 31 mins ago
esote
2,93111038
2,93111038
asked 2 hours ago
Accountant مAccountant م
1827
1827
2
$begingroup$
I'd be really interested to see what shutdown(char* msg) does.
$endgroup$
– pacmaninbw
1 hour ago
$begingroup$
In the use case that was provided, since you can effectively predict the size, I would think it would be more natural to have a string buffer and the size passed in instead of creating it dynamically.
$endgroup$
– Neil Edelman
28 mins ago
add a comment |
2
$begingroup$
I'd be really interested to see what shutdown(char* msg) does.
$endgroup$
– pacmaninbw
1 hour ago
$begingroup$
In the use case that was provided, since you can effectively predict the size, I would think it would be more natural to have a string buffer and the size passed in instead of creating it dynamically.
$endgroup$
– Neil Edelman
28 mins ago
2
2
$begingroup$
I'd be really interested to see what shutdown(char* msg) does.
$endgroup$
– pacmaninbw
1 hour ago
$begingroup$
I'd be really interested to see what shutdown(char* msg) does.
$endgroup$
– pacmaninbw
1 hour ago
$begingroup$
In the use case that was provided, since you can effectively predict the size, I would think it would be more natural to have a string buffer and the size passed in instead of creating it dynamically.
$endgroup$
– Neil Edelman
28 mins ago
$begingroup$
In the use case that was provided, since you can effectively predict the size, I would think it would be more natural to have a string buffer and the size passed in instead of creating it dynamically.
$endgroup$
– Neil Edelman
28 mins ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Formatting
Most C formatting guides do not include spaces around the arguments to function calls, nor the expressions within an if-statement. For an example of a C style most C programmers would find acceptable, see OpenBSD's style(9)
manual.
I choose to associate *
with the variable name, rather than floating between the type and name. This disambiguates the following example:
int *a, b;
Here, a
is a pointer to an integer, but b
is only an integer. By moving the asterisk next to the name, it makes this clearer.
int length = strlen ( str );
char * newStr = malloc (length * 2 );
if ( !newStr) shutDown ( "can't allocate memory" ) ;
Becomes:
int const len = strlen(str);
char *const new_str = malloc(len * 2);
if (new_str == NULL) {
shutDown("can't allocate memory");
}
Error checking
Rather than calling shutDown()
and exit()
ing the program, you should instead return an error value which can be checked by the caller of str_to_hex()
. Because you return a pointer, you can return NULL
to indicate an error occurred and the caller should check errno
.
Likewise, on some systems your program can incorrectly exit when length == 0
. If we look at the manual page for malloc(3)
:
Return Value
The malloc() and calloc() functions return a pointer to the allocated memory that is suitably aligned for any kind of variable. On error, these functions return NULL. NULL may also be returned by a successful call to malloc() with a size of zero, or by a successful call to calloc() with nmemb or size equal to zero.
So by returning NULL
we account for the case where malloc(3)
returns NULL on success.
if (new_str == NULL) {
shutDown("can't alloc memory");
}
Becomes:
if (new_str == NULL) {
return NULL;
}
If you choose, you can also check if str
is NULL before calling strlen()
. This is up to you, and it's not uncommon in C to ignore this case and leave it as user error.
Looping
Use the size_t
type in your loop rather than int
. size_t
is guaranteed be wide enough to hold any array index, while int
is not.
Using i
rather than x
is more common for looping variables.
The y
variable isn't needed. You can simply use str[i]
in its place.
In terms of performance there's likely a faster option than using sprintf()
. You should look into strtol(3)
.
Conclusion
Here is the code I ended up with:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char *
str_to_hex(char const *const str)
{
size_t const len = strlen(str);
char *const new_str = malloc(len * 2);
if (new_str == NULL) {
return NULL;
}
for (size_t i = 0; i < len; ++i) {
sprintf(new_str + i * 2, "%02X", str[i]);
}
return new_str;
}
int
main(void)
{
char *str = "abz";
char *hex = str_to_hex(str);
if (hex == NULL && strlen(str) != 0) {
/* error ... */
}
printf("%sn",hex);
free(hex);
}
Hope this helps!
$endgroup$
add a comment |
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Formatting
Most C formatting guides do not include spaces around the arguments to function calls, nor the expressions within an if-statement. For an example of a C style most C programmers would find acceptable, see OpenBSD's style(9)
manual.
I choose to associate *
with the variable name, rather than floating between the type and name. This disambiguates the following example:
int *a, b;
Here, a
is a pointer to an integer, but b
is only an integer. By moving the asterisk next to the name, it makes this clearer.
int length = strlen ( str );
char * newStr = malloc (length * 2 );
if ( !newStr) shutDown ( "can't allocate memory" ) ;
Becomes:
int const len = strlen(str);
char *const new_str = malloc(len * 2);
if (new_str == NULL) {
shutDown("can't allocate memory");
}
Error checking
Rather than calling shutDown()
and exit()
ing the program, you should instead return an error value which can be checked by the caller of str_to_hex()
. Because you return a pointer, you can return NULL
to indicate an error occurred and the caller should check errno
.
Likewise, on some systems your program can incorrectly exit when length == 0
. If we look at the manual page for malloc(3)
:
Return Value
The malloc() and calloc() functions return a pointer to the allocated memory that is suitably aligned for any kind of variable. On error, these functions return NULL. NULL may also be returned by a successful call to malloc() with a size of zero, or by a successful call to calloc() with nmemb or size equal to zero.
So by returning NULL
we account for the case where malloc(3)
returns NULL on success.
if (new_str == NULL) {
shutDown("can't alloc memory");
}
Becomes:
if (new_str == NULL) {
return NULL;
}
If you choose, you can also check if str
is NULL before calling strlen()
. This is up to you, and it's not uncommon in C to ignore this case and leave it as user error.
Looping
Use the size_t
type in your loop rather than int
. size_t
is guaranteed be wide enough to hold any array index, while int
is not.
Using i
rather than x
is more common for looping variables.
The y
variable isn't needed. You can simply use str[i]
in its place.
In terms of performance there's likely a faster option than using sprintf()
. You should look into strtol(3)
.
Conclusion
Here is the code I ended up with:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char *
str_to_hex(char const *const str)
{
size_t const len = strlen(str);
char *const new_str = malloc(len * 2);
if (new_str == NULL) {
return NULL;
}
for (size_t i = 0; i < len; ++i) {
sprintf(new_str + i * 2, "%02X", str[i]);
}
return new_str;
}
int
main(void)
{
char *str = "abz";
char *hex = str_to_hex(str);
if (hex == NULL && strlen(str) != 0) {
/* error ... */
}
printf("%sn",hex);
free(hex);
}
Hope this helps!
$endgroup$
add a comment |
$begingroup$
Formatting
Most C formatting guides do not include spaces around the arguments to function calls, nor the expressions within an if-statement. For an example of a C style most C programmers would find acceptable, see OpenBSD's style(9)
manual.
I choose to associate *
with the variable name, rather than floating between the type and name. This disambiguates the following example:
int *a, b;
Here, a
is a pointer to an integer, but b
is only an integer. By moving the asterisk next to the name, it makes this clearer.
int length = strlen ( str );
char * newStr = malloc (length * 2 );
if ( !newStr) shutDown ( "can't allocate memory" ) ;
Becomes:
int const len = strlen(str);
char *const new_str = malloc(len * 2);
if (new_str == NULL) {
shutDown("can't allocate memory");
}
Error checking
Rather than calling shutDown()
and exit()
ing the program, you should instead return an error value which can be checked by the caller of str_to_hex()
. Because you return a pointer, you can return NULL
to indicate an error occurred and the caller should check errno
.
Likewise, on some systems your program can incorrectly exit when length == 0
. If we look at the manual page for malloc(3)
:
Return Value
The malloc() and calloc() functions return a pointer to the allocated memory that is suitably aligned for any kind of variable. On error, these functions return NULL. NULL may also be returned by a successful call to malloc() with a size of zero, or by a successful call to calloc() with nmemb or size equal to zero.
So by returning NULL
we account for the case where malloc(3)
returns NULL on success.
if (new_str == NULL) {
shutDown("can't alloc memory");
}
Becomes:
if (new_str == NULL) {
return NULL;
}
If you choose, you can also check if str
is NULL before calling strlen()
. This is up to you, and it's not uncommon in C to ignore this case and leave it as user error.
Looping
Use the size_t
type in your loop rather than int
. size_t
is guaranteed be wide enough to hold any array index, while int
is not.
Using i
rather than x
is more common for looping variables.
The y
variable isn't needed. You can simply use str[i]
in its place.
In terms of performance there's likely a faster option than using sprintf()
. You should look into strtol(3)
.
Conclusion
Here is the code I ended up with:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char *
str_to_hex(char const *const str)
{
size_t const len = strlen(str);
char *const new_str = malloc(len * 2);
if (new_str == NULL) {
return NULL;
}
for (size_t i = 0; i < len; ++i) {
sprintf(new_str + i * 2, "%02X", str[i]);
}
return new_str;
}
int
main(void)
{
char *str = "abz";
char *hex = str_to_hex(str);
if (hex == NULL && strlen(str) != 0) {
/* error ... */
}
printf("%sn",hex);
free(hex);
}
Hope this helps!
$endgroup$
add a comment |
$begingroup$
Formatting
Most C formatting guides do not include spaces around the arguments to function calls, nor the expressions within an if-statement. For an example of a C style most C programmers would find acceptable, see OpenBSD's style(9)
manual.
I choose to associate *
with the variable name, rather than floating between the type and name. This disambiguates the following example:
int *a, b;
Here, a
is a pointer to an integer, but b
is only an integer. By moving the asterisk next to the name, it makes this clearer.
int length = strlen ( str );
char * newStr = malloc (length * 2 );
if ( !newStr) shutDown ( "can't allocate memory" ) ;
Becomes:
int const len = strlen(str);
char *const new_str = malloc(len * 2);
if (new_str == NULL) {
shutDown("can't allocate memory");
}
Error checking
Rather than calling shutDown()
and exit()
ing the program, you should instead return an error value which can be checked by the caller of str_to_hex()
. Because you return a pointer, you can return NULL
to indicate an error occurred and the caller should check errno
.
Likewise, on some systems your program can incorrectly exit when length == 0
. If we look at the manual page for malloc(3)
:
Return Value
The malloc() and calloc() functions return a pointer to the allocated memory that is suitably aligned for any kind of variable. On error, these functions return NULL. NULL may also be returned by a successful call to malloc() with a size of zero, or by a successful call to calloc() with nmemb or size equal to zero.
So by returning NULL
we account for the case where malloc(3)
returns NULL on success.
if (new_str == NULL) {
shutDown("can't alloc memory");
}
Becomes:
if (new_str == NULL) {
return NULL;
}
If you choose, you can also check if str
is NULL before calling strlen()
. This is up to you, and it's not uncommon in C to ignore this case and leave it as user error.
Looping
Use the size_t
type in your loop rather than int
. size_t
is guaranteed be wide enough to hold any array index, while int
is not.
Using i
rather than x
is more common for looping variables.
The y
variable isn't needed. You can simply use str[i]
in its place.
In terms of performance there's likely a faster option than using sprintf()
. You should look into strtol(3)
.
Conclusion
Here is the code I ended up with:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char *
str_to_hex(char const *const str)
{
size_t const len = strlen(str);
char *const new_str = malloc(len * 2);
if (new_str == NULL) {
return NULL;
}
for (size_t i = 0; i < len; ++i) {
sprintf(new_str + i * 2, "%02X", str[i]);
}
return new_str;
}
int
main(void)
{
char *str = "abz";
char *hex = str_to_hex(str);
if (hex == NULL && strlen(str) != 0) {
/* error ... */
}
printf("%sn",hex);
free(hex);
}
Hope this helps!
$endgroup$
Formatting
Most C formatting guides do not include spaces around the arguments to function calls, nor the expressions within an if-statement. For an example of a C style most C programmers would find acceptable, see OpenBSD's style(9)
manual.
I choose to associate *
with the variable name, rather than floating between the type and name. This disambiguates the following example:
int *a, b;
Here, a
is a pointer to an integer, but b
is only an integer. By moving the asterisk next to the name, it makes this clearer.
int length = strlen ( str );
char * newStr = malloc (length * 2 );
if ( !newStr) shutDown ( "can't allocate memory" ) ;
Becomes:
int const len = strlen(str);
char *const new_str = malloc(len * 2);
if (new_str == NULL) {
shutDown("can't allocate memory");
}
Error checking
Rather than calling shutDown()
and exit()
ing the program, you should instead return an error value which can be checked by the caller of str_to_hex()
. Because you return a pointer, you can return NULL
to indicate an error occurred and the caller should check errno
.
Likewise, on some systems your program can incorrectly exit when length == 0
. If we look at the manual page for malloc(3)
:
Return Value
The malloc() and calloc() functions return a pointer to the allocated memory that is suitably aligned for any kind of variable. On error, these functions return NULL. NULL may also be returned by a successful call to malloc() with a size of zero, or by a successful call to calloc() with nmemb or size equal to zero.
So by returning NULL
we account for the case where malloc(3)
returns NULL on success.
if (new_str == NULL) {
shutDown("can't alloc memory");
}
Becomes:
if (new_str == NULL) {
return NULL;
}
If you choose, you can also check if str
is NULL before calling strlen()
. This is up to you, and it's not uncommon in C to ignore this case and leave it as user error.
Looping
Use the size_t
type in your loop rather than int
. size_t
is guaranteed be wide enough to hold any array index, while int
is not.
Using i
rather than x
is more common for looping variables.
The y
variable isn't needed. You can simply use str[i]
in its place.
In terms of performance there's likely a faster option than using sprintf()
. You should look into strtol(3)
.
Conclusion
Here is the code I ended up with:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char *
str_to_hex(char const *const str)
{
size_t const len = strlen(str);
char *const new_str = malloc(len * 2);
if (new_str == NULL) {
return NULL;
}
for (size_t i = 0; i < len; ++i) {
sprintf(new_str + i * 2, "%02X", str[i]);
}
return new_str;
}
int
main(void)
{
char *str = "abz";
char *hex = str_to_hex(str);
if (hex == NULL && strlen(str) != 0) {
/* error ... */
}
printf("%sn",hex);
free(hex);
}
Hope this helps!
edited 20 mins ago
answered 35 mins ago
esoteesote
2,93111038
2,93111038
add a comment |
add a comment |
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2
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I'd be really interested to see what shutdown(char* msg) does.
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– pacmaninbw
1 hour ago
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In the use case that was provided, since you can effectively predict the size, I would think it would be more natural to have a string buffer and the size passed in instead of creating it dynamically.
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– Neil Edelman
28 mins ago