How can I convert an integer to a hexadecimal string in C?

How can I convert an integer to a hexadecimal string in C?

Example: The integer 50 would be converted to the hexadecimal string "32" or "0x32".

7 Answers

This code

int a = 5; printf("%x\n", a); 

prints

5 

This code

int a = 5; printf("0x%x\n", a); 

prints

0x5 

This code

int a = 89778116; printf("%x\n", a); 

prints

559e7c4 

If you capitalize the x in the format it capitalizes the hex value:

int a = 89778116; printf("%X\n", a); 

prints

559E7C4 

If you want to print pointers you use the p format specifier:

char* str = "foo"; printf("0x%p\n", str); 

prints

0x01275744 

The following code takes an integer and makes a string out of it in hex format:

int num = 32424; char hex[5]; sprintf(hex, "%x", num); puts(hex); 

gives

7ea8 

Usually with printf (or one of its cousins) using the %x format specifier.

Interesting that these answers utilize printf like it is a given. printf converts the integer to a Hexadecimal string value.

//************************************************************* // void prntnum(unsigned long n, int base, char sign, char *outbuf) // unsigned long num = number to be printed // int base = number base for conversion; decimal=10,hex=16 // char sign = signed or unsigned output // char *outbuf = buffer to hold the output number //************************************************************* void prntnum(unsigned long n, int base, char sign, char *outbuf) { int i = 12; int j = 0; do{ outbuf[i] = "0123456789ABCDEF"[num % base]; i--; n = num/base; }while( num > 0); if(sign != ' '){ outbuf[0] = sign; ++j; } while( ++i < 13){ outbuf[j++] = outbuf[i]; } outbuf[j] = 0; } 

I made a librairy to make Hexadecimal / Decimal conversion without the use of stdio.h. Very simple to use :

char* dechex (int dec); 

This will use calloc() to to return a pointer to an hexadecimal string, this way the quantity of memory used is optimized, so don't forget to use free()

Here the link on github :

To convert an integer to a string also involves char array or memory management.

To handle that part for such short arrays, code could use a compound literal, since C99, to create array space, on the fly. The string is valid until the end of the block.

#define UNS_HEX_STR_SIZE ((sizeof (unsigned)*CHAR_BIT + 3)/4 + 1) // compound literal v--------------------------v #define U2HS(x) unsigned_to_hex_string((x), (char[UNS_HEX_STR_SIZE]) {0}, UNS_HEX_STR_SIZE) char *unsigned_to_hex_string(unsigned x, char *dest, size_t size) { snprintf(dest, size, "%X", x); return dest; } int main(void) { // 3 array are formed v v v printf("%s %s %s\n", U2HS(UINT_MAX), U2HS(0), U2HS(0x12345678)); char *hs = U2HS(rand()); puts(hs); // `hs` is valid until the end of the block } 

Output

FFFFFFFF 0 12345678 5851F42D 

This answer is for those, who need to start from string in decimal representation (not from int).

1. Convert your string representation of the number to an integer value (you can use int atoi( const char * str ); function
2. Once you have your integer you can print it as HEX using, for example, sprintf function with %x as a format parameter and you integer as a value parameter

Here is a working example:

#include <stdio.h> int main(void) { int n; char hex_val[50]; n = atoi("100663296"); sprintf(hex_val, "%x", n); printf("%s", hex_val); return 0; }