Question

我想创建一个将unsigned char转换为unsigned int并将其存储到数组中的函数。但是，最终会出现错误

从不兼容的指针类型传递'sprintf'的参数1。

int main(void) {
    unsigned char key[16] = "1234567812345678";
    phex(key, 16); //store into an array here
}

uint64_t* phex(unsigned char* string, long len)
{
    uint64_t hex[len];
    int count = 0;

    for(int i = 0; i < len; ++i) {
        count = i * 2;
        sprintf(hex + count, "%.2x", string[i]);
    }

    for(int i = 0; i < 32; i++)
        printf(hex[i]);

    return hex;
}

Answer 1

正如评论已经说过，你的代码有问题...... 首先，sprintf函数与您希望/期望它完全相反。接下来，在函数中创建一个局部变量，并返回指向它的指针。一旦函数退出，指针就无效了。我看到的第三个问题是你永远不会将回报值分配给任何东西......

关于如何修复代码的提议：

unsigned* phex(unsigned char* string, long len);

int main(void) {
    int i;
    unsigned char key[16] = "1234567812345678";

    unsigned* ints = phex(key,16); //store into an array here

    for(i = 0; i < 16; i++)
        printf("%d ", ints[i]);

    //never forget to deallocate memory
    free(ints);

    return 0;
}

unsigned* phex(unsigned char* string, long len)
{
    int i;
    //allocate memory for your array
    unsigned* hex = (unsigned*)malloc(sizeof(unsigned) * len);

    for(i = 0; i < len; ++i) {
        //do char to int conversion on every element of char array
        hex[i] = string[i] - '0';
    }

    //return integer array
    return hex;
}

将unsigned char（array）转换为unsigned int（array）

1 个答案: