Question

下面是我在课堂上做作业的代码段，但是当我尝试编译它时会弹出：

[Error] invalid conversion from 'void*' to 'int*' [-fpermissive]

指针1-3上的

，选项和两个选项。我仍然是新手程序员，我不确定为什么现在会发生这种情况。

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

int main() 
{
    int *pointer1, *pointer2, *pointer3;
    int *choice;

    char * option;
    char * option1;

    pointer1 = malloc ( sizeof(int) );
    pointer2 = malloc ( sizeof(int) );
    pointer3 = malloc ( sizeof(int) );
    choice   = malloc ( sizeof(int) );
    option = malloc ( sizeof(char) );
    option1 = malloc ( sizeof(char) );
}

Answer 1

Malloc函数将返回void*。您需要将其转换为正确的类型，如下所示：

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


int main()
{
    int* pointer1, * pointer2, * pointer3;
    int* choice;
    char* option;
    char* option1;

    pointer1 = (int*) malloc(sizeof(int));

    pointer2 = (int*) malloc(sizeof(int));

    pointer3 = (int*) malloc(sizeof(int));

    choice = (int*) malloc(sizeof(int));

    option = (char*) malloc(sizeof(char));

    option1 = (char*) malloc(sizeof(char));

    // Clean up after yourself!!!
    free(pointer1);
    free(pointer2);
    free(pointer3);
    free(choice);
    free(option);
    free(option1);

    return 0;
}

而且，不要忘记释放已用的内存！

编译时出现奇怪的错误

1 个答案: