警告：从不兼容的指针类型赋值 [默认启用] 并从不同大小的整数转换为指针 [-Wint-to-pointer-cast]

Question

当我尝试运行我的代码时，出现警告，我不知道为什么....请帮助我，这是我的代码：

#include<stdio.h>
#include<stdlib.h>
#include<sys/time.h>

#define TESTTIMES 2000


void test1()
{
    // printf("--------TEST1----------\n");
    int i = 0;
    for (i = 0; i < 150; i++) {
        char *p = (char*)MyMalloc(1);
        MyFree(p);
    }
}


void test2()
{
    // printf("--------TEST2----------\n");
    int *a[150], i;
    for (i = 0; i < 150; i++) {
        char *p = (char*)MyMalloc(1);
        a[i] = p;
        if ((i + 1) % 50 == 0) {
            int j = i + 1  - 50;
            while (j <= i) {
                MyFree(a[j]);
                j += 1;
            }
        }
    }
}

void test3()
{
    // printf("--------TEST3----------\n");
    int flag, cnt = 0, top = 0;
    char *b[50];

    srand(time(NULL));
    while(cnt < 50) {
        flag = rand() % 2;
        if (flag) {
            char *p = (char*)MyMalloc(1);
            b[top] = p;
            top++;
            cnt += 1;
        } else {
            if (top > 0) {
                top -= 1;
                MyFree(b[top]);
            }
        }
    }
    while (top > 0) {
        top--;
        MyFree(b[top]);
    }
}

void test4() 
{
    // printf("--------TEST4----------\n");
    int flag, size, cnt = 0, top = 0;
    char *b[50];

    srand(time(NULL));
    while(cnt < 50) {
        flag = rand() % 2;
        size = rand() % 64 + 1;
        if (flag) {
            char *p = (char*)MyMalloc(size);
            b[top] = p;
            top++;
            cnt += 1;
        } else {
            if (top > 0) {
                top -= 1;
                MyFree(b[top]);
            }
        }
    }
    while (top > 0) {
        top--;
        MyFree(b[top]);
    }
}

void test5()
{
    // printf("--------TEST5----------\n");
    int i = 0;
    while (i < TESTTIMES) {
        char *p = MyMalloc(4096);
        MyFree(p);
        i++;
    }
}

void test6()
{
    // printf("--------TEST6----------\n");

    int top = 0, i = 0;
    char *b[100];

    while(1) {
        char *p = (char*)MyMalloc(100); 
        if (p) {
            b[top] = p;
            top++;
        } else {
            break;
        }
    }

    while(i < top){
        MyFree(b[i]);
        i++;
    }
}

我收到以下警告：

ind.c: In function \u2018test1\u2019:
ind.c:79:19: warning: cast to pointer from integer of different size [-Wint-to-pointer-cast]
         char *p = (char*)MyMalloc(1);
                   ^
ind.c: In function \u2018test2\u2019:
ind.c:90:19: warning: cast to pointer from integer of different size [-Wint-to-pointer-cast]
         char *p = (char*)MyMalloc(1);
                   ^
ind.c:91:14: warning: assignment from incompatible pointer type [enabled by default]
         a[i] = p;
              ^
ind.c: In function \u2018test3\u2019:
ind.c:112:23: warning: cast to pointer from integer of different size [-Wint-to-pointer-cast]
             char *p = (char*)MyMalloc(1);
                       ^
ind.c: In function \u2018test4\u2019:
ind.c:140:23: warning: cast to pointer from integer of different size [-Wint-to-pointer-cast]
             char *p = (char*)MyMalloc(size);
                       ^
ind.c: In function \u2018test5\u2019:
ind.c:162:19: warning: initialization makes pointer from integer without a cast [enabled by default]
         char *p = MyMalloc(4096);
                   ^
ind.c: In function \u2018test6\u2019:
ind.c:176:19: warning: cast to pointer from integer of different size [-Wint-to-pointer-cast]
         char *p = (char*)MyMalloc(100); 
                   ^

是不是因为 GCC 版本不同？我该如何解决？这是C编程，我收到了两个警告，例如警告：从不兼容的指针类型赋值[默认启用]和从不同大小的整数转换为指针[-Wint-to-pointer-cast]。

Answer 1

您收到这些错误的原因：

警告：从不同大小的整数转换为指针

是因为您的主文件不知道MyMalloc 是如何定义的，所以它默认为返回int 的函数。您可以通过将 #include "mymalloc.h" 放在主文件顶部来解决此问题。

如果您这样做，您会看到另一个错误，特别是 buffer 和 memory 被定义了两次。这是因为您在头文件中定义了这两个变量。正因为如此，变量不应该在头文件中定义。

由于您只在 mymalloc.c 中使用 buffer 和 memory，您应该将它们移动到该文件中。这样可以避免多重定义错误。出于同样的原因，您也应该移动struct node。