用printf打印字符串十六进制格式，扭曲的结果 [英] use printf to print character string in hex format, distorted results

问题描述

我要打印字符串十六进制格式，

I want to print character string in hex format,

在机器A，像

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

在机器B

 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");

种子变化中1，2 3 ....

seed varies among 1, 2 3....

字节代funcion是：

the bytes generation funcion is:

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;
}

但我得到的结果从printf的，如：

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 

为什么有那么多的 FFFFF 为 mesg_check ？

有没有这种现象任何潜在的原因是什么？
谢谢！

are there any potential reasons for this phenomenon? thanks!

推荐答案

下面是说明我的问题想的你可能有一个小程序：

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（在其纯字符签署），我得到这样的输出：

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

  0 10 7f ffffff80 ffffffff

值 255 ，当存储在一个（签字）字符，存储为 1 。在的printf 通话，它的晋升（签字） INT - 但％ 2X格式告诉的printf 把它当作一个 unsigned int类型，所以它显示 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.

请确保您的 MESG 和 mesg_check 数组被定义为无符号的字符数组，而不是普通的字符。

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

更新：超过一年后重读这个答案，我意识到这是不完全正确。下面是我的系统上正常工作的程序，并且将任何合理的系统上几乎可以肯定工作：

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;
}

输出是：

 00 10 7f 80 ff

类型的参数 unsigned char型被提升为（签字） INT （假设 INT 可容纳类型的所有值 unsigned char型，即 INT_MAX＆GT; = UCHAR_MAX ，它是在几乎所有的系统的情况下）。因此，参数改编[I] 提升为 INT ，而％02X 格式需要类型的参数 unsigned int类型。

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，但不的非常的状态直接对应符号​​和无符号类型的参数是可以互换的，只要它们是两种类型的范围之内 - 这是这里的情况。

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.

要成为的完全的正确，你需要确保的说法实际上是类型 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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