使用printf打印十六进制格式的字符串，结果失真 [英] Use printf to print character string in hexadecimal format, distorted results

问题描述

我想在机器 A 上以十六进制格式打印一个字符串.例如:

I want to print a character string in hexadecimal format on machine A. Something like:

ori_mesg = gen_rdm_bytestream (1400, seed)
sendto(machine B, ori_mesg, len(mesg))

在机器 B 上

recvfrom(machine A, mesg)

mesg_check = gen_rdm_bytestream (1400, seed)

for(i=0;i<20;i++){
    printf("%02x ", *(mesg+i)& 0xFF);
}
printf("\n");
for(i=0; i<20; i++){
    printf("%02x ", *(mesg_check+i));
}
printf("\n");

seed 在 1, 2, 3, ...

seed varies among 1, 2, 3, ...

字节生成函数为:

u_char *gen_rdm_bytestream (size_t num_bytes, unsigned int seed)
{
    u_char *stream = malloc (num_bytes+4);
    size_t i;

    u_int16_t seq = seed;
    seq = htons(seq);
    u_int16_t tail = num_bytes;
    tail = htons(tail);
    memcpy(stream, &seq, sizeof(seq));
    srand(seed);
    for (i = 3; i < num_bytes+2; i++){
        stream[i] = rand ();
    }
    memcpy(stream+num_bytes+2, &tail, sizeof(tail));

    return stream;
}

但我从 printf 得到的结果如下:

But I got results from printf like:

00 01 00 67 c6 69 73 51 ff 4a ec 29 cd ba ab f2 fb e3 46 7c
00 01 00 67 ffffffc6 69 73 51 ffffffff 4a ffffffec 29 ffffffcd ffffffba ffffffab fffffff2 fffffffb ffffffe3 46 7c

或

00 02 88 fa 7f 44 4f d5 d2 00 2d 29 4b 96 c3 4d c5 7d 29 7e
00 02 00 fffffffa 7f 44 4f ffffffd5 ffffffd2 00 2d 29 4b ffffff96 ffffffc3 4d ffffffc5 7d 29 7e

为什么mesg_check有这么多fffff?

这种现象是否有任何潜在的原因?

Are there any potential reasons for this phenomenon?

推荐答案

这是一个小程序，它说明了我认为您可能遇到的问题:

Here's a small program that illustrates the problem I think you might be having:

#include <stdio.h>
int main(void) {
    char arr[] = { 0, 16, 127, 128, 255 };
    for (int i = 0; i < sizeof arr; i ++) {
        printf(" %2x", arr[i]);
    }
    putchar('\n');
    return 0;
}

在我的系统上(对纯 char 进行签名)，我得到以下输出:

On my system (on which plain char is signed), I get this output:

  0 10 7f ffffff80 ffffffff

值 255 存储在(有符号的)char 中时，存储为 -1.在 printf 调用中，它被提升为(签名)int -- 但 "%2x" 格式告诉 printf> 将其视为 unsigned int，因此它显示 fffffffff.

The value 255, when stored in a (signed) char, is stored as -1. In the printf call, it's promoted to (signed) int -- but the "%2x" format tells printf to treat it as an unsigned int, so it displays fffffffff.

确保您的 mesg 和 mesg_check 数组被定义为 unsigned char 数组，而不是普通的 char.

Make sure that your mesg and mesg_check arrays are defined as arrays of unsigned char, not plain char.

更新: 一年多后重读这个答案，我意识到它不太正确.这是一个可以在我的系统上正常运行的程序，并且几乎肯定可以在任何合理的系统上运行:

UPDATE: Rereading this answer more than a year later, I realize it's not quite correct. Here's a program that works correctly on my system, and will almost certainly work on any reasonable system:

#include <stdio.h>
int main(void) {
    unsigned char arr[] = { 0, 16, 127, 128, 255 };
    for (int i = 0; i < sizeof arr; i ++) {
        printf(" %02x", arr[i]);
    }
    putchar('\n');
    return 0;
}

输出为:

 00 10 7f 80 ff

unsigned char 类型的参数被提升为(有符号)int(假设 int 可以保存 类型的所有值unsigned char，即 INT_MAX >= UCHAR_MAX，实际上所有系统都是这种情况).所以参数 arr[i] 被提升为 int，而 " %02x" 格式需要一个 unsigned int 类型的参数.

An argument of type unsigned char is promoted to (signed) int (assuming that int can hold all values of type unsigned char, i.e., INT_MAX >= UCHAR_MAX, which is the case on practically all systems). So the argument arr[i] is promoted to int, while the " %02x" format requires an argument of type unsigned int.

C 标准强烈暗示，但没有完全直接声明，相应的有符号和无符号类型的参数可以互换，只要它们在两种类型的范围内 - 即案例在这里.

The C standard strongly implies, but doesn't quite state directly, that arguments of corresponding signed and unsigned types are interchangeable as long as they're within the range of both types -- which is the case here.

要完全正确，您需要确保参数实际上是 unsigned int 类型:

To be completely correct, you need to ensure that the argument is actually of type unsigned int:

printf("%02x", (unsigned)arr[i]);

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