c - 使用Calloc将十进制数组转换为二进制字符串数组

我留下了一些代码，但基本上我有一个小数数组，我试图将它转换成二进制数，但作为一个字符串数组。对C来说是全新的，现在真的是在抓救命稻草。我很不确定malloc和calloc是如何使用的。这是我试图让它对第一个数字/二进制字符串起作用的尝试：

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

void binstringconvert (unsigned long int *decimals, char **binarystrings);


int main (int argc, char **argv)
{
    // initialize variables
    int numLines = 9;
    int binaryLength = 32;
    unsigned long int decimals[numLines];
    decimals[0] = 3241580200;

    // convert decimal array to 32-bit binary string array
    char **binarystrings = calloc (numLines, binaryLength);
    binstringconvert(decimals, binarystrings);

    // test print
    printf("\n\n%lu in binary number system is: ", decimals[0]);
    printf("\n%s", binarystrings[0]);
}

void binstringconvert (unsigned long int *decimals, char **binarystrings)
{
    int c, k;

    for (c = 31; c >= 0; c--)
    {
        k = decimals[0] >> c;
        if (k & 1)
            binarystrings[0][c] = '1';
        else
            binarystrings[0][c] = '0';
    }
}

我是否正确初始化了binarystrings？我能像以前那样给每个人写信吗？现在它给了我一个错误。

最佳答案

很抱歉，我解释说您想向函数传递多个数字，并将它们全部转换为二进制字符串并返回。不管怎样，这个例子仍然有效。在这两种情况下，要将32位数字放入字符串，需要33个字符（nul终止字符为1）。另外，您正在以相反的顺序编写二进制字符串。
只要稍微调整一下就可以纠正顺序。例子：

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

enum { DWRD = 32 };

void binstringconvert (unsigned *decimals, char (*binarystrings)[DWRD+1], int n);

int main (void)
{
    /* initialize variables */
    unsigned int decimals[] = { 1, 255, 65535, 8388607,
                                3241580200, 2898560974,
                                4294967295, 3097295382,
                                1076482445, 1234567890 };
    char (*binarystrings)[DWRD+1] = {NULL};
    int i, n = sizeof decimals/sizeof *decimals;

    binarystrings = calloc (n, sizeof *binarystrings);
    binstringconvert (decimals, binarystrings, n);

    /* test print */
    for (i = 0; i < n; i++)
        printf (" %10u : %s\n", decimals[i], binarystrings[i]);

    free (binarystrings);

    return 0;
}

void binstringconvert (unsigned *decimals, char (*binarystrings)[DWRD+1], int n)
{
    int c, i, k;
    for (i = 0; i < n; i++) {
        for (c = 31; c >= 0; c--)
        {
            k = decimals[i] >> c;
            if (k & 1)
                binarystrings[i][31-c] = '1';
            else
                binarystrings[i][31-c] = '0';
        }
        binarystrings[i][DWRD] = 0;  /* nul-terminate */
    }
}

输出

$ ./bin/binstrings
          1 : 00000000000000000000000000000001
        255 : 00000000000000000000000011111111
      65535 : 00000000000000001111111111111111
    8388607 : 00000000011111111111111111111111
 3241580200 : 11000001001101101001011010101000
 2898560974 : 10101100110001001000011111001110
 4294967295 : 11111111111111111111111111111111
 3097295382 : 10111000100111001111101000010110
 1076482445 : 01000000001010011101000110001101
 1234567890 : 01001001100101100000001011010010