c - Directly assigning values to C Pointers

Question

I've just started learning C and I've been running some simple programs using MinGW for Windows to understand how pointers work. I tried the following:

#include <stdio.h>

int main(){
    int *ptr;
    *ptr = 20;
    printf("%d", *ptr);
    return 0;
}

which compiled properly but when I run the executable it doesn't work - the value isn't printed to the command line, instead I get an error message that says the .exe file has stopped working.

However when I tried storing the value in an int variable and assign *ptr to the memory address of that variable as shown below:

#include <stdio.h>

int main(){
    int *ptr;
    int q = 50;
    ptr = &q;
    printf("%d", *ptr);
    return 0;
}

it works fine.

My question is, why am I unable to directly set a literal value to the pointer? I've looked at tutorials online for pointers and most of them do it the same way as the second example.

Any help is appreciated.

score 83 · Accepted Answer

问题是您没有初始化指针。你已经创建了一个指向“任何你想要的地方”的指针——它可能是一些其他变量的地址，或者你的代码中间，或者一些根本没有映射的内存。

您需要int在内存中的某处创建一个变量int *以指向该变量。

您的第二个示例执行此操作，但它执行与此处无关的其他操作。这是您需要做的最简单的事情：

int main(){
    int variable;
    int *ptr = &variable;
    *ptr = 20;
    printf("%d", *ptr);
    return 0;
}

在这里，int变量没有被初始化——但这很好，因为你只是要用20. 关键是指针被初始化为指向variable. 事实上，如果你愿意，你可以分配一些原始内存来指向：

int main(){
    void *memory = malloc(sizeof(int));
    int *ptr = (int *)memory;
    *ptr = 20;
    printf("%d", *ptr);
    free(memory);
    return 0;
}

score 27 · Accepted Answer

第一个带有评论的程序

#include <stdio.h>

int main(){
    int *ptr;             //Create a pointer that points to random memory address

    *ptr = 20;            //Dereference that pointer, 
                          // and assign a value to random memory address.
                          //Depending on external (not inside your program) state
                          // this will either crash or SILENTLY CORRUPT another 
                          // data structure in your program.  

    printf("%d", *ptr);   //Print contents of same random memory address
                          // May or may not crash, depending on who owns this address

    return 0;             
}

带评论的第二个程序

#include <stdio.h>

int main(){
    int *ptr;              //Create pointer to random memory address

    int q = 50;            //Create local variable with contents int 50

    ptr = &q;              //Update address targeted by above created pointer to point
                           // to local variable your program properly created

    printf("%d", *ptr);    //Happily print the contents of said local variable (q)
    return 0;
}

关键是你不能使用指针，除非你知道它被分配给你自己管理的地址，要么将它指向你创建的另一个变量，要么指向 malloc 调用的结果。

之前使用它会创建依赖于未初始化内存的代码，这在最好的情况下会崩溃，但在最坏的情况下有时会工作，因为随机内存地址恰好位于程序已经拥有的内存空间内。如果它覆盖了您在程序中其他地方使用的数据结构，上帝会帮助您。

score 3 · Accepted Answer

在第一个示例中，ptr 尚未初始化，因此它指向一个未指定的内存位置。当您将某些内容分配给这个未指定的位置时，您的程序就会崩溃。

在第二个示例中，当您说 ptr = &q 时设置了地址，所以您没问题。

score -1 · Accepted Answer

您可以为指针设置一个值，但是一旦您使用“new”为它请求内存。这就是您的代码的外观

int main(){
    int *ptr;
    ptr = new int;    //ask for memory
    *ptr = 20;
    printf("%d", *ptr);
    return 0;
}

c - Directly assigning values to C Pointers

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