使用 malloc 时从 `void*' 到 `char*' 的无效转换?

2022-01-23 00:00:00 malloc g++ c++

我在下面的代码中遇到问题，第 5 行出现错误:

I'm having trouble with the code below with the error on line 5:

错误:从 void* 到 char*

我正在使用带有代码块的 g++，并尝试将此文件编译为 cpp 文件.有关系吗?

I'm using g++ with codeblocks and I tried to compile this file as a cpp file. Does it matter?

#include <openssl/crypto.h>
int main()
{
    char *foo = malloc(1);
    if (!foo) {
        printf("malloc()");
        exit(1);
    }
    OPENSSL_cleanse(foo, 1);
    printf("cleaned one byte
");
    OPENSSL_cleanse(foo, 0);
    printf("cleaned zero bytes
");
}

推荐答案

在C++中，需要强制转换malloc()的返回

In C++, you need to cast the return of malloc()

char *foo = (char*)malloc(1);

相关文章