c - 为什么这个 *ptr 不给出存储在 ptr 变量中包含的内存地址处的实际值？

Question

这是一个关于指针的基本 C 程序:

#include <stdio.h>

int main() {
    int variable = 20;
    int *pointerToVariable;

    pointerToVariable = &variable;

    printf("Address of variable: %x\n", &variable);
    printf("Address of variable: %x\n", pointerToVariable);
    printf("Address of variable: %x\n", *pointerToVariable); // * is the DEREFERENCING OPERATOR/INDIRECTION OPERATOR in C. 
                                                                //It gives the value stored at the memory address 
                                                                //stored in the variable which is its operand.
    getchar();
    return 0;
}

这会产生以下输出:

Address of variable: 8ffbe4
Address of variable: 8ffbe4
Address of variable: 14

但是*pointerToVariable应该打印20，不是吗？因为 * 给出了存储在其操作数中存储的内存地址处的实际值，对吗？我错过了什么？

Answer 1

14 是 20 的 HEX 值。

将 printf 格式说明符更改为 %d 而不是 %x，以将 20 作为输出

printf("Address of variable: %d\n", *pointerToVariable);

此外，指针的正确格式说明符是 %p，因此

printf("Address of variable: %x\n", pointerToVariable);

必须是

printf("Address of variable: %p\n", (void *)pointerToVariable);