赋值从整数生成指针，不需要强制转换和其他问题

Question

int32result[pos] = returnedInteger;

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


    int charToInt32(char * clearText, unsigned int * int32result[])
    {
        int i = 0, j = 0, pos = 0;          /* Counters */
        int dec[4] = {24, 16, 8, 0};        /* Byte positions in an array*/
        unsigned int returnedInteger = 0;           /*   so we can change the order*/ 

        for (i=0; clearText[i]; i+=4) {
            returnedInteger = 0;
            for (j=0; j <= 3 ; j++) {
                returnedInteger |= (unsigned int) (clearText[i+j] << dec[j]) ;
            }

            int32result[pos] = returnedInteger;
            pos++;
        }
        return pos;
    }



    int main()
    {
        int i = 0;
        unsigned int * int32result[1024] = {0};
        char * clearText =  "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

        printf(">>Clear: %s\n", clearText);
        charToInt32(clearText, int32result); // Do the conversion to int32.
        printf(">>Int32 converted: ");
        for (i=0; int32result[i]; i++)
            printf("%u ", (unsigned int) int32result[i]);
        printf("\n");
        return 0;
    }

printf("%u ", (unsigned int) int32result[i])

谢谢你的帮助。

Answer 1

unsigned int * int32result[1024]

声明由1024个无符号int指针组成的数组。我想您应该在这里使用一个数组

unsigned int int32result[1024]

在chartoint32中也有类似的问题，其中unsigned int*int32result[]参数指定了一个unsigned int数组的数组。您只有一个数组，因此可以传递无符号int*int32result（即删除[])。

那么您的其余代码就可以正常工作了。像这样的呼叫

charToInt32(clearText, int32result);

charToInt32(clearText, &int32result[0]);