警告：格式'％d'期望类型为'int *'的参数，但是参数3的类型为'uint8_t * {aka unsigned char *}] >> [

Question

我遇到的问题是：

warning: format ‘%d’ expects argument of type ‘int *’, but argument 3 has type ‘uint8_t * {aka unsigned char *}

而且我不确定为什么会这样，因为我认为int和uint8_t是可互换的。

这是我的代码：

uint8_t** fileRead(char* file, int* pointer) {
    FILE* file = fopen(file, "r");
    int count = 0;
    pointer = &count;
    fscanf(file, "%d", &count); //this retrieves a single integer
    uint8_t** result = (uint8_t**) malloc(*pointer * sizeof(uint8_t*));
    while (file != NULL) {
        for (uint8_t i = 0; i < *pointer; i++) {
            uint8_t* a = (uint8_t*) malloc(sizeof(uint8_t));
            uint8_t* b = (uint8_t*) malloc(sizeof(uint8_t));
            uint8_t* c = (uint8_t*) malloc(sizeof(uint8_t));
            fscanf(file, "%d", a);
            fscanf(file, "%d", b);
            fscanf(file, "%d", c);
            if (a == NULL || b == NULL || c == NULL) {
                uint8_t* npointer = NULL;
                fclose(file);
                free(result);
                return NULL;
            } else {
                result[i][0] = *a;
                result[i][1] = *b;
                result[i][2] = *c;
                free(a);
                free(b);
                free(c);
           }
        }
    }
    return result; 
}

并且错误发生在以下几行：

fscanf(file, "%d", a);
fscanf(file, "%d", b);
fscanf(file, "%d", c);

非常感谢您

我一直遇到这样的问题：警告：格式'％d'期望类型为'int *'的参数，但是参数3的类型为'uint8_t * {aka unsigned char *}，我不确定为什么会这样发生的原因是...

Answer 1

我认为int和uint8_t可以互换。

Answer 2

Answer 3

Answer 4