C 程序仅由于格式说明符而显示错误

Question

这个问题已经有答案了:

Correct format specifier to print pointer or address? (6 个回答)

What happens when I use the wrong format specifier? (2 个回答)

已关闭 4 年前。

由于我在 for 循环中打印数组地址时使用的格式说明符，该程序显示错误。如果我使用除 %p 之外的任何格式说明符来打印地址，例如 %ld、%u、%d 等，则会显示如下错误:

prog.c: In function 'main':
prog.c:18:10: warning: format '%ld' expects argument of type 'long int', but argument 3 has type 'int *' [-Wformat=]
   printf("Address arr[%d] is %ld\n", i, &arr[i]); 
          ^

这是我的代码:

 // C program to demonstrate that array elements are stored 
// contiguous locations 

#include <stdio.h> 
int main() 
{ 
    // an array of 10 integers. If arr[0] is stored at 
    // address x, then arr[1] is stored at x + sizeof(int) 
    // arr[2] is stored at x + sizeof(int) + sizeof(int) 
    // and so on. 
    int arr[5], i; 

    printf("Size of integer in this compiler is %lu\n", sizeof(int)); 

    for (i = 0; i < 5; i++) 
        // The use of '&' before a variable name, yields 
        // address of variable. 
        printf("Address arr[%d] is %ld\n", i, &arr[i]); 

    return 0;

我无法理解为什么在这里使用 %p 以及我可以使用什么来使我的程序没有错误

Answer 1

该标准提供了“%p”用于打印指针地址，并且它期望提供的类型为void*。您可以重新编写代码以正确使用“%p”，如下所示:

#include <stdio.h> 

int main (void) 
{ 
    int arr[5], i; 

    printf ("Size of integer in this compiler is %lu\n", sizeof(int)); 

    for (i = 0; i < 5; i++) 
        printf ("Address arr[%d] is %p\n", i, (void*)&arr[i]); 

    return 0;
}

示例使用/输出

$ ./bin/ptraddrarr
Size of integer in this compiler is 4
Address arr[0] is 0x7ffe251cac50
Address arr[1] is 0x7ffe251cac54
Address arr[2] is 0x7ffe251cac58
Address arr[3] is 0x7ffe251cac5c
Address arr[4] is 0x7ffe251cac60

不推荐，但您可以使用“0x%lx”格式说明符来获得相同的输出，例如

        printf ("Address arr[%d] is 0x%lx\n", i, (unsigned long)&arr[i]); 

但是，不要使用为此目的明确提供的正确格式说明符。