为指向函数内部结构的指针分配内存 [英] Allocate memory for a pointer to a struct inside a function

问题描述

我正在尝试编写一个单独的程序，该程序调用一个函数为一定数量的学生"结构动态分配内存.

I am trying to write a separate program that calls a function to dynamically allocate memory for a certain number of "student" structs.

我的主程序太大而无法处理，所以我创建了一个较小的程序来帮助我更轻松地弄清楚我在做什么:

My main program was too large to mess with so I created a smaller program to help me more easily figure out what I'm doing:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include "memtest.h"

void structMemInit(myName **);

int main(void){

    myName *myNameIs;

    structMemInit(&myNameIs);

    printf("%s\n", myNameIs[1].name);

    return 0;
}

void structMemInit(myName **myNameIs){
    *myNameIs = (myName *) calloc(5, sizeof(myName));
    if(myNameIs == NULL){
        printf("allocating memory didn't work!\n");
        exit(1);
    }
    else if(myNameIs != NULL)
        (*myNameIs)[1].name = "Zach";
}

memtest.h 文件是:

the memtest.h file is:

typedef struct{
    char *name;
}myName;

以上程序试图做的就是将指向结构的指针传递给函数structMemInit"并为该结构分配干净的空间.然后为了查看它是否有效，我将变量 char 设置为一个名称.在这一切发生之后，您离开函数并返回主函数.然后我在结构中打印名称以显示它有效.

All the above program is trying to do is pass a pointer to a struct into the function "structMemInit" and allocate clean space for the struct. Then to see show that it worked, I am setting the variable char to a name. After all this happens, you leave the function and go back to main. Then I print the name in the struct to show it worked.

运行时，程序总是给出段错误.

When run, the program always gives a segfault.

如果你想知道为什么我有一个单独的 .h 文件，我的实际程序要大得多，并且有几个全局声明的结构，我的老师要求我为这些结构保留一个单独的 .h.

In case you are wondering why I have a separate .h file, my actual program is much larger and has several globally declared structs and I am required by my instructor to keep a separate .h for those structs.

谢谢！扎克

推荐答案

如果你坚持传递指针，应该是通过引用，所以它是一个指向指针的指针，因为函数itslef会修改指针中的地址.

If you insist on passing the pointer, it should be by reference, so it's a pointer to a pointer, as the function itslef will modify the address in the pointer.

如果你的函数返回一个指针，你会好得多.

You're much better off if your function returns a pointer.

//function for allocating the memory
myName *structMemInit(void)
{
    myName *myNameIs = (myName *)calloc(1, sizeof(myName));
    if (myNameIs == NULL) {
        printf("allocating memory didn't work!\n");
        exit(1);
    } else {                     // don't repeat the negation of the condition
        myNameIs->name = "Zach"; // No need for \0, it's automatic
        return myNameIs;
    }
}

//Usage
myName *myNameIs = structMemInit();

顺便说一句，它是 int main(void) 而不是 int main().

BTW it's int main(void) not int main().

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