-1
#include<stdio.h>
#include<stdlib.h>
int *a;
int b;
int main()
{
b = 5;
*a = &b;
printf("%x \n",a);
printf("%d",*a);
printf("\n %d",b);
return 0;
}
In code output this error appears :
ponteiros.c: In function ‘main’: ponteiros.c:13:8: warning: assignment to ‘int’ from ‘int *’ makes integer from pointer without a cast [-Wint-conversion] *a = &b;
1
The formatter for a pointer is the p and not the x:
#include <stdio.h>
int main() {
int b = 5;
int *a = &b;
printf("%p\n", a);
printf("%d\n", *a);
printf("%d", b);
}
Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.
Browser other questions tagged c pointer
You are not signed in. Login or sign up in order to post.
Did the answer solve your question? Do you think you can accept it? See [tour] if you don’t know how you do it. This would help a lot to indicate that the solution was useful to you. You can also vote on any question or answer you find useful on the entire site.
– Maniero