c - 从 C 中不兼容的指针类型错误获取赋值

Question

我必须编写一个代码来声明和初始化三个变量:double、int 和 string(5 位，包括空字符)，然后使用 printf 和 sizeof 打印大小地址和值

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

int main() {

    int number = 5, *pI;
    pI = &number; // Assigning Integer variable to integer pointer
    double number2 = 10.5, *pD;
    pD = &number2; // Assigning Double variable to Double pointer
    char arry[] = "dogs", *pc;
    pc = &arry; // Assigning Char Array variable to Char Array Pointer

    // Size, Address and Value stored in each Integer Variable, Double Variable and Character Variable
    printf("Integer Size is = %d , Address is = 0x%p and Value is = %d: \n",
            sizeof(number), &number, number);
    fflush(stdout);
    printf("Double Size is = %d, Address is = 0x%p and Value is = %lf: \n",
            sizeof(number2), &number2, number2);
    fflush(stdout);
    printf(
            "Character Array Size is = %d, Address is = 0x%p and Value is = %s: \n",
            sizeof(arry), &arry, arry);
    fflush(stdout);

    // Pointers Size, Address and Value stored in each Integer Pointer, Double Pointer and Charecter pointer
    printf(
            "Pointer Integer Size is = %d , Address is = 0x%p and Value is = %d: \n",
            sizeof(pI), &pI, *pI);
    fflush(stdout);
    printf(
            "Pointer Double Size is = %d, Address is = 0x%p and Value is = %lf: \n",
            sizeof(pD), &pD, *pD);
    fflush(stdout);
    printf(
            "Pointer Character Array Size is = %d, Address is = 0x%p and Value is = %s: \n",
            sizeof(pc), &pc, pc);
    fflush(stdout);

    return 0;
}

Answer 1

char arry[] = "dogs", *pc;
pc = &arry;

arry 的类型为 char[5]，它是 char 的数组，因此 &arry 是指向 char 的指针和数组。由于 pc 的类型为 char *，因此指针不兼容。

pc = arry;

这里 arry 衰减为指向其第一个元素的指针，并且此赋值是有效的。

另一方面，如果您确实想要一个指向 char 数组的指针(如注释中所写)，则必须像这样声明它:

char (*pc)[5];

此外，您还应该在 printf 中使用正确的标识符，并将指针强制转换为 void *。对于size_t，您应该使用%zu。