malloc(0) 的行为 [英] behaviour of malloc(0)

问题描述

int main()
{
    char *p;
    p = (char* ) malloc(sizeof(char) * 0);
    printf("Hello Enter the data without spaces :
");
    scanf("%s",p);
    printf("The entered string is %s
",p);
    //puts(p);
}

在编译并运行上面的代码后，即使我们为指针 p 分配了一个 0 字节的内存，程序也能够读取字符串.

On compiling the above code and running it , the program is able to read the string even though we assigned a 0 byte memory to the pointer p .

语句 p = (char* ) malloc(0) 中实际发生了什么?

What actually happens in the statement p = (char* ) malloc(0) ?

推荐答案

malloc() 将返回什么是实现定义的，但使用该指针是未定义的行为.未定义行为意味着任何事情都可能发生，从程序无故障运行到崩溃，所有安全赌注都已关闭.

It is implementation defined what malloc() will return but it is undefined behavior to use that pointer. And Undefined behavior means that anything can happen literally from program working without glitch to a crash, all safe bets are off.

C99 标准:

7.22.3 内存管理功能
第 1 段:

如果请求的空间大小为零，则行为是实现定义的:要么返回空指针，要么行为就像大小是某个非零值，除了返回的指针不得用于访问一个对象.

If the size of the space requested is zero, the behavior is implementation-defined: either a null pointer is returned, or the behavior is as if the size were some nonzero value, except that the returned pointer shall not be used to access an object.

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