Question

在下面的代码中，应该从注释中不言自明，而不是打印字符串tmp，它实际上打印了argv[2]的值。我已通过LLDB运行它，一切正常，所有媒体间值都正确，tmp在调用long2hex后保持正确的字符串，但它没有打印到屏幕上。我可以通过LLDB，检查变量等等，但不知道如何调试这样的东西，即使经过一些广泛的谷歌搜索。有没有人知道为什么要打印argv[2]，或者可以指示我使用LLDB调试行为？

正在使用两个命令行参数argv[1]="1c0111001f010100061a024b53535009181c"和argv[2]="686974207468652062756c6c277320657965"

运行程序

如果有人有兴趣，这是matasano加密挑战之一！设置1，问题2.可能有一种更简单的方法，但这就是我想出来的！

以下是代码：

#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <math.h>

/* Converts a character to the corresponding integer value */
uint8_t char2hex (const char a_char) {
    uint8_t rv = 0;

    if (a_char >= 48 && a_char <= 57) /* '0'-'9' */
        rv = (uint8_t)a_char - 48;
    else if (a_char >= 65 && a_char <=70) /* 'A'-'F' */
        rv = (uint8_t)a_char - 55;
    else if (a_char >= 97 && a_char <= 102) /* 'a'-'f' */
        rv = (uint8_t)a_char - 87;

    return rv;
}

/* Converts an integer value to its corresponding hex character */
char hex2char (uint8_t a_hex) {
    char rv;

    if (a_hex >= 0 && a_hex <= 9)
        rv = a_hex + 48;
    else if (a_hex >= 10 && a_hex <= 15)
        rv = a_hex + 87;
    else
        rv = 'z';

    return rv;

}

/* Coverts an unsigned 32-bit integer value into its hexadecimal representation */
void long2hex (uint32_t val, char result[9]) {
    char hex[9];

    int j = 0;
    for (int i = 0; i < 4; i++,j+=2){
        uint8_t tmpVal;
        tmpVal = (val>>((3-i)*8));
        hex[j] = hex2char((tmpVal >> 4) & 0x0F);
        hex[j+1] = hex2char(tmpVal & 0x0F);
    }
    hex[8] = '\0';

    result = hex;
}


/* Converts an 8 character hex string into its unsigned integer form */
uint32_t hex2int32(uint8_t h[8]) {
    uint32_t b = 0;

    for (int i = 0; i < 8; i++)
        b += (h[i]) << (4*(7-i));

    return b;

}

/* Splits a string into 8 character chunks */
char ** split_hex_8 (const char * hex_string, int * num_strings) {
    int string_length = strlen(hex_string);
    int num_blocks = (int)ceil((double)string_length / 8);

    char ** split_hex = (char **) malloc(num_blocks * sizeof(char*));

    int length_counter = string_length;

    for (int i = 0; i < num_blocks; i++) {
        split_hex[i] = (char *) malloc(9);
        if (length_counter < 8) {
            memcpy(split_hex[i], hex_string+(i*8), length_counter);
            for (int j=length_counter; j < 8; j++)
                (split_hex[i])[j] = '0';
        } else
            memcpy(split_hex[i], hex_string+(i*8), 8);
        (split_hex[i])[8] = '\0';
    }

    *num_strings = num_blocks;

    return split_hex;
}



int main (int argc, char * argv[]) {

    char * hex_str_1 = argv[1];
    char * hex_str_2 = argv[2];

    int len1 = strlen(argv[1]);
    int len2 = strlen(argv[2]);

    char ** hex_split_1, ** hex_split_2;

    int num_blocks_1, num_blocks_2;

    /* Split the input strings */
    hex_split_1 = split_hex_8(hex_str_1,&num_blocks_1);
    hex_split_2 = split_hex_8(hex_str_2,&num_blocks_2);


    /* For each of the inputs, xor them together by xoring the individual 32-bit unsigned integers together */
    for (int i = 0; i < num_blocks_1; i++) {
        uint32_t val1, val2, xor;
        uint8_t hex1[8] = {0};
        uint8_t hex2[8] = {0};
        for (int j = 0; j < 8; j++){ /* convert each hex character in each string to its integer equivalent */
            hex1[j] = char2hex((hex_split_1[i])[j]);
            hex2[j] = char2hex((hex_split_2[i])[j]);
        }
        free(hex_split_1[i]); /* We are now done with this memory, so free it. */
        free(hex_split_2[i]);

        /* Shift in all of the 8-bit decoded hex into a 32-bit integer - gets the unsigned 32 bit representation of each 8 byte block of input */
        val1 = hex2int32(hex1);
        val2 = hex2int32(hex2);

        /* Create some temporary storage for the new hex string */
        char * tmp;
        tmp = malloc(9);

        xor = val1^val2;
        /* convert this xor-ed value to it's corresponding hex representation, store it in tmp */
        long2hex(xor,tmp);
        /* print tmp to the screen */
        /* THIS IS THE ERROR: instead of printing tmp to the screen, it ends up printing out argv[2]. */
        printf("%s",tmp);
        /* done with tmp! */
        free(tmp);
    }


    /* don't forget to free the containers for the split hex */
    free(hex_split_1);
    free(hex_split_2);


    return 0;

}

Answer 1

result = hex;

那是你的问题。 result是一个局部变量。分配给它不会传播回调用者。你想要的：

strncpy(result, hex, sizeof(result));

程序打印完全不同的字符串到它应该

1 个答案: