C语言 如何在C中将整数转换为十六进制字符串？
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How can I convert an integer to a hexadecimal string in C?
提问by user417894
How can I convert an integer to a hexadecimal string in C?
如何在C中将整数转换为十六进制字符串？
Example: The integer 50would be converted to the hexadecimal string "32"or "0x32".
示例：整数50将被转换为十六进制字符串"32"或"0x32".
回答by C Johnson
This code
这段代码
int a = 5;
printf("%x\n", a);
prints
印刷
5
This code
这段代码
int a = 5;
printf("0x%x\n", a);
prints
印刷
0x5
This code
这段代码
int a = 89778116;
printf("%x\n", a);
prints
印刷
559e7c4
If you capitalize the x in the format it capitalizes the hex value:
如果在格式中将 x 大写，则它会将十六进制值大写：
int a = 89778116;
printf("%X\n", a);
prints
印刷
559E7C4
If you want to print pointers you use the p format specifier:
如果要打印指针，请使用 p 格式说明符：
char* str = "foo";
printf("0x%p\n", str);
prints
印刷
0x01275744
回答by Pithikos
The following code takes an integerand makes a stringout of it in hex format:
以下代码接受一个整数并以十六进制格式将其生成一个字符串：
int num = 32424;
char hex[5];
sprintf(hex, "%x", num);
puts(hex);
gives
给
7ea8
回答by Jerry Coffin
Usually with printf(or one of its cousins) using the %xformat specifier.
通常与printf（或其表亲之一）使用%x格式说明符。
回答by dlb
Interesting that these answers utilize printflike it is a given.
printfconverts the integer to a Hexadecimal string value.
有趣的是，这些答案printf就像给定的一样使用。
printf将整数转换为十六进制字符串值。
//*************************************************************
// 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;
}
回答by kevmuret
I made a librairy to make Hexadecimal / Decimal conversion without the use of stdio.h. Very simple to use :
我制作了一个库来进行十六进制/十进制转换而不使用stdio.h. 使用非常简单：
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()
这将用于calloc()返回一个指向十六进制字符串的指针，这样优化了使用的内存量，所以不要忘记使用free()
Here the link on github : https://github.com/kevmuret/libhex/
这里是 github 上的链接：https: //github.com/kevmuret/libhex/
回答by chux - Reinstate Monica
To convert an integer to a stringalso involves chararray or memory management.
将整数转换为字符串还涉及char数组或内存管理。
To handle that part for such short arrays, code could use a compound literal, since C99, to create array space, on the fly. The stringis valid until the end of the block.
为了处理这种短数组的那部分，代码可以使用复合文字，从 C99 开始，即时创建数组空间。该字符串在块结束之前一直有效。
#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
输出
FFFFFFFF 0 12345678
5851F42D
