指针不会工作的printf（） [英] Pointer will not work in printf()

问题描述

有一个问题与打印指针移出。每次我尝试和编译程序下面我得到以下错误：

  pointers.c：11：警告：格式'％P'需要键入'无效*'，但参数2的类型'诠释*'
 

我显然简单的东西在这里，但我见过类似code等examles，这应该是工作。

这里的code，任何帮助将是巨大的！

 的#include＆LT;＆stdio.h中GT;    INT主要（无效）
    {
       INT X = 99;
       INT * PT1;       PT1 =安培; X;       的printf（价值在P1数：％d \\ n，* PT1）;
       的printf（P1地址：％P \\ N，PT1）;       返回0;
    }
 

解决方案

只需将您的INT指针转换为void之一：

 的printf（P1地址：％P \\ N（无效*）PT1）;
 

您code是安全的，但你是用 -Wformat 警告标志进行编译，将键入检查的printf的调用（ ）和 scanf（）的。

Having an issue with printing a pointer out. Every time I try and compile the program below i get the following error:

pointers.c:11: warning: format ‘%p’ expects type ‘void *’, but argument 2 has type ‘int *’

I'm obviously missing something simple here, but from other examles of similar code that I have seen, this should be working.

Here's the code, any help would be great!

#include <stdio.h>

    int main(void)
    {
       int x = 99;
       int *pt1;

       pt1 = &x;

       printf("Value at p1: %d\n", *pt1);
       printf("Address of p1: %p\n", pt1);

       return 0;
    }

解决方案

Simply cast your int pointer to a void one:

printf( "Address of p1: %p\n", ( void * )pt1 );

Your code is safe, but you are compiling with the -Wformat warning flag, that will type check the calls to printf() and scanf().

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