Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
315 views
in Technique[技术] by (71.8m points)

c - Pointer will not work in printf()

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

       return 0;
    }
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Simply cast your int pointer to a void one:

printf( "Address of p1: %p
", ( 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().


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...