我正在测试如何使用带有指针算法的数组名称来访问数组元素,并且我想出了以下程序:
#include <stdio.h>
int main(){
// definition of array using designators
int a[2][2][2] = {
[0] = {[0] = {1, 2}, [1] = {3, 4}},
[1] = {[0] = {5, 6}, [1] = {7, 8}}
};
printf("7th element (pointer): %p\n", *(*(a + 1) + 1) + 0);
printf("8th element (pointer): %p\n", *(*(a + 1) + 1) + 1);
return 0;
}
尽管程序可以正常工作,并且可以正确打印所有内容:
7th element (pointer): 0x7ffd4b09a1c8
8th element (pointer): 0x7ffd4b09a1d0
我在编译时收到警告,对于在%p
内使用printf()
占位符的每一行都这样说:
warning: format ‘%p’ expects argument of type ‘void *’, but argument 2 has type ‘int *’ [-Wformat=]
printf("7th element (pointer): %p\n", *(*(a + 1) + 1) + 0);
~^ ~~~~~~~~~~~~~~~~~~~
%ls
因此,乍一看,我似乎只需要将指针转换为(void *)
,如果我这样做并更改行即可:
printf("8th element (pointer): %p\n", *(*(a + 1) + 1) + 1);
此行:
printf("8th element (pointer): %p\n", (void *)(*(a + 1) + 1) + 1);
我收到警告:
warning: pointer of type ‘void *’ used in arithmetic [-Wpointer-arith]
printf("8th element (pointer): %p\n", (void *)(*(a + 1) + 1) + 1);
^
并且编译器为第8个元素计算一个不同的地址,该地址仅比elemnet的地址大1个字节:
7th element (pointer): 0x7ffd9e5e1bf8
8th element (pointer): 0x7ffd9e5e1bf9
我也试图像这样修复它(增加了一个大括号):
printf("8th element (pointer): %p\n", (void *)((*(a + 1) + 1) + 1));
警告消失了,但地址的计算方式仍然不同:
7th element (pointer): 0x7ffca6c6c468
8th element (pointer): 0x7ffca6c6c470
看起来编译器需要指针类型来计算地址,如果我强制转换,它将不会核心地计算地址。有谁知道我该怎么做才能删除警告并以正确的方式计算地址?
答案 0 :(得分:1)
如果您需要在int*
上进行算术运算,并且将值视为void*
,则在执行算术运算后将值强制转换为:< / p>
printf("7th element (pointer): %p\n", (void *)((int*)*(*(a + 1) + 1) + 0));
printf("8th element (pointer): %p\n", (void *)((int*)*(*(a + 1) + 1) + 1));
答案 1 :(得分:0)
您可以将指针存储到一个单独的变量中,并将其传递给printf:
void *ptr7th = (*(*(a + 1) + 1) + 0);
void *ptr8th = (*(*(a + 1) + 1) + 1);
printf("7th element (pointer): %p\n", ptr7th);
printf("8th element (pointer): %p\n", ptr8th);
将gcc版本9.3.0与-Wall
参数一起使用不会返回任何警告。