打印函数中定义的指针的值和地址？

Question

我认为这是一个很容易编码的东西,但我在C语法中遇到了问题,我只是用C++编程.

#include <stdio.h>
#include <stdlib.h>

void pointerFuncA(int* iptr){
/*Print the value pointed to by iptr*/
printf("Value:  %x\n", &iptr );

/*Print the address pointed to by iptr*/

/*Print the address of iptr itself*/
}

int main(){

void pointerFuncA(int* iptr); 

return 0;
}

显然这段代码只是一个骨架,但我想知道如何在函数和主要工作之间进行通信,以及打印指向的地址和iptr本身的语法？由于函数无效,如何将所有三个值发送到main？

我认为地址是这样的:

printf("Address of iptr variable: %x\n", &iptr );

我知道这是一个简单的问题,但我在网上发现的所有例子都得到了价值,但它在主要定义为

int iptr = 0;

我需要创建一些任意值吗？

谢谢!

Answer 1

阅读评论

#include <stdio.h>
#include <stdlib.h>

void pointerFuncA(int* iptr){
/*Print the value pointed to by iptr*/
printf("Value:  %d\n", *iptr );

/*Print the address pointed to by iptr*/
printf("Value:  %p\n", iptr );

/*Print the address of iptr itself*/
printf("Value:  %p\n", &iptr );
}

int main(){
int i = 1234; //Create a variable to get the address of
int* foo = &i; //Get the address of the variable named i and pass it to the integer pointer named foo
pointerFuncA(foo); //Pass foo to the function. See I removed void here because we are not declaring a function, but calling it.

return 0;
}

输出:

Value:  1234
Value:  0xffe2ac6c
Value:  0xffe2ac44

Answer 2

要访问指针指向的值,必须使用间接运算符*.

要打印指针本身,只需访问没有运算符的指针变量.

要获取指针变量的地址,请使用&运算符.

void pointerFuncA(int* iptr){
    /*Print the value pointed to by iptr*/
    printf("Value:  %x\n", *iptr );

    /*Print the address pointed to by iptr*/
    printf("Address of value: %p\n", (void*)iptr);

    /*Print the address of iptr itself*/
    printf("Address of iptr: %p\n", (void*)&iptr);
}

该%p格式的操作需要相应的说法是void*,这样的三分球投射到这种类型的有必要.