c - fprintf 不会将以零结尾的十六进制数正确写入文件

Question

只要“标签”的初始值为奇数，我就能够将正确的十六进制值写入输出文件。但我想以十六进制形式为偶数和奇数初始“标签”值编写输出。如果'tag'的初始值是偶数且大于数字 20，并且我尝试将十六进制数写入输出文件，我只会得到字符 ちㄠ 而不是 A0 1。但是，如果我将其更改为以十进制格式写入我得到正确的值。如果我只将 fieldnumber[0] 写入文件而不是将 fieldnumber[0] 和 'secondaryvalue' 写入文件，那么我将得到正确的十六进制输出值。此外，如果我将 fieldnumber[0] 更改为仅写入十进制输出并将“secondaryvalue”保留为十六进制，它将给我写入值。我的代码如下(省略无关行:

int uint32_pack (uint8_t *fieldnumber, uint32_t value, uint8_t *out);

int main(){

uint32_t initvalue = 2;
int return_rv;
uint8_t *tag = (uint8_t *) malloc(sizeof(uint8_t));

uint8_t *tempout= (uint8_t *) malloc(sizeof(uint32_t));
*tag = 20; //even number that when processed won't write the correct value

return_rv = uint32_pack (tag, initvalue, tempout);

free(tempout);

    }    

/* === pack() === */
/* Pack an unsigned 32-bit integer in base-128 encoding, and return the number
 of bytes needed: this will be 5 or less. */

int uint32_pack (uint8_t *fieldnumber, uint32_t value, uint8_t *out)
{
  unsigned rv = 0;
  FILE *outfile;
  FILE *wiretypetag;
  int secondaryvalue;
  outfile = fopen("hexdata.txt","w");
  wiretypetag = fopen("wiretype.txt","w");


    //encodes wire type and the field number
    if (*fieldnumber <16){
       *fieldnumber <<= 3;
       fprintf(wiretypetag,"%x",*fieldnumber);
       }
    if (*fieldnumber < 32){
       *fieldnumber <<= 3;
       secondaryvalue = 0x01;
       fprintf(wiretypetag,"%x %x",fieldnumber[0],secondaryvalue);
           }
    if (*fieldnumber < 48){
        *fieldnumber += 0x10;
        *fieldnumber &= 0x1F;
        *fieldnumber <<= 3;
        secondaryvalue = 0x02;
        fprintf(wiretypetag,"%x %x",fieldnumber[0], secondaryvalue);
        }          

  /* assert: value<128 */   
    out[rv++] = value;

    fclose(outfile);
    fclose(wiretypetag);

    return rv;
}

Answer 1

filednumber[] 是一个 long long 值(8 字节)，需要以 %llx 格式打印，而不是 %x在执行此操作时(查看手册页)，您可能需要不同的输出格式，例如 %lx 或 %Lx，具体取决于您使用的编译器和 clib。