C语言 如何在C中打印变量地址？
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How to print variable addresses in C?
提问by nambvarun
When i run this code.
当我运行此代码时。
#include <stdio.h>
void moo(int a, int *b);
int main()
{
int x;
int *y;
x = 1;
y = &x;
printf("Address of x = %d, value of x = %d\n", &x, x);
printf("Address of y = &d, value of y = %d, value of *y = %d\n", &y, y, *y);
moo(9, y);
}
void moo(int a, int *b)
{
printf("Address of a = %d, value of a = %d\n", &a, a);
printf("Address of b = %d, value of b = %d, value of *b = %d\n", &b, b, *b);
}
I keep getting this error in my compiler.
我在编译器中不断收到此错误。
/Volumes/MY USB/C Programming/Practice/addresses.c:16: warning: format ‘%d' expects type ‘int', but argument 2 has type ‘int *'
/Volumes/MY USB/C Programming/Practice/addresses.c:17: warning: format ‘%d' expects type ‘int', but argument 2 has type ‘int **'
/Volumes/MY USB/C Programming/Practice/addresses.c:17: warning: format ‘%d' expects type ‘int', but argument 3 has type ‘int *'
/Volumes/MY USB/C Programming/Practice/addresses.c: In function ‘moo':
/Volumes/MY USB/C Programming/Practice/addresses.c:23: warning: format ‘%d' expects type ‘int', but argument 2 has type ‘int *'
/Volumes/MY USB/C Programming/Practice/addresses.c:24: warning: format ‘%d' expects type ‘int', but argument 2 has type ‘int **'
/Volumes/MY USB/C Programming/Practice/addresses.c:24: warning: format ‘%d' expects type ‘int', but argument 3 has type ‘int *'
Could you help me?
你可以帮帮我吗？
Thanks
谢谢
blargman
吹牛者
回答by Carl Norum
You want to use %pto print a pointer. From the spec:
你想用来%p打印一个指针。从规范：
pThe argument shall be a pointer tovoid. The value of the pointer is converted to a sequence of printing characters, in an implementation-defined manner.
p参数应是指向 的指针void。指针的值以实现定义的方式转换为打印字符序列。
And don't forget the cast, e.g.
并且不要忘记演员阵容，例如
printf("%p\n",(void*)&a);
回答by Ron Nuni
When you intend to print the memory address of any variable or a pointer, using %dwon't do the job and will cause some compilation errors, because you're trying to print out a number instead of an address, and even if it does work, you'd have an intent error, because a memory address is not a number. the value 0xbfc0d878is surely not a number, but an address.
当您打算打印任何变量或指针的内存地址时，使用%d不会完成这项工作并会导致一些编译错误，因为您试图打印出数字而不是地址，即使它确实有效，您会遇到意图错误，因为内存地址不是数字。该值0xbfc0d878肯定不是数字，而是地址。
What you should use is %p. e.g.,
你应该使用的是%p. 例如，
#include<stdio.h>
int main(void) {
int a;
a = 5;
printf("The memory address of a is: %p\n", (void*) &a);
return 0;
}
Good luck!
祝你好运！
回答by Amarendra Deo
To print the address of a variable, you need to use the %pformat. %dis for signed integers. For example:
要打印变量的地址，您需要使用%p格式。%d用于有符号整数。例如：
#include<stdio.h>
void main(void)
{
int a;
printf("Address is %p:",&a);
}
回答by Praveer Kumar
I tried in online compiler https://www.onlinegdb.com/online_c++_compiler
我试过在线编译器 https://www.onlinegdb.com/online_c++_compiler
int main()
{
cout<<"Hello World";
int x = 10;
int *p = &x;
printf("\nAddress of x is %p\n", &x); // 0x7ffc7df0ea54
printf("Address of p is %p\n", p); // 0x7ffc7df0ea54
return 0;
}
回答by skaz
Looks like you use %p: Print Pointers
看起来你使用 %p: Print Pointers
