Question

我一直试图调用此函数，但我一直收到错误＆＃34;标识符未找到＆＃34; （是的，我知道我的变量名称不是最实用的。）

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

typedef struct _POOL
{
    int size;
    void* memory;

} Pool;

int main()
{
    printf("Enter the number of bytes you want to allocate\n");
    int x = getchar();
    allocatePool(x);

    return 0;
}

Pool* allocatePool(int x)
{
    Pool* p = (Pool*)malloc(x);
    return p;
}

Answer 1

也许您想要做的事情应该是这样的：

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

typedef struct pool
{
    int size;
    void* memory;
} pool;

pool allocatePool(int x);

int main (int argc, char *argv[])
{
    int x = 0;
    pool *p = NULL;
    printf("enter the number of bytes you want to allocate//>\n");
    if (scanf("%d", &x) == 1 && x > 0 && (p = allocatePool(x)) != NULL)
    {
        // Do something using p
        // ...
        // Treatment is done, now freeing memory
        free(p->memory);

        free(p);
        p = NULL; // Not useful right before exiting, but most often good practice.
        return 0;
    }
    else
    {
        // Show a message error or do an alternative treatment
        return 1;
    }
}

pool allocatePool(int x)
{
    pool *p = malloc(sizeof(pool));
    if (p != NULL)
    {
        p->memory = malloc(x);
        p->size = (p->memory == NULL) ? 0 : x;
    }
    return p;
}

Answer 2

我认为你的程序应该是这样的：

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

typedef struct _Pool
{
    int size;
    void* memory;
} Pool;

Pool* allocatePool(int x)
{
    static Pool p;//no pointer here you also need memory for your integer or you have to allocate memory for the integer too
    p.size=x;
    p.memory=malloc(x);//just allocate memory for your void pointer
    return &p;//return the adress of the Pool
}

int main()
{
    printf("enter the number of bytes you want to allocate//>\n");
    int x;
    scanf("%d", &x);//if you use getchar() for reading you pass the ascii value of the number not the normal number
    allocatePool(x);
    return 0;
}

试图在c中分配记忆功能？

2 个答案: