在没有return语句的情况下检查函数的返回值 [英] Checking return value of a function without return statement

问题描述

请解释为什么有时有时不需要return语句?

Please explain why sometimes return statement is not needed?

函数具有返回类型，但是缺少返回语句.同时，程序可以编译并正常工作.

Function has a return type, but return statement missing. Meanwhile, program compiles and works fine.

请帮助我更好地理解这一点

Please help me understand this better

char* handleInput() {
    fgets(buffer, 1024, stdin);
//    return buffer;       <---- COMMENTED RETURN
}

void main() {
        char* ptr = handleInput();
        int flag = atoi(ptr);    
        if (flag < 0) break;    
        printf("You entered: %s\n", ptr);
}

推荐答案

基本上返回的是愚蠢的运气.您将获得返回时CPU寄存器中的信息.例如，如果返回的值是AX，而 char * 恰好是AX，则您很幸运.我相信这是未定义的行为；也就是说，C语言规范没有告诉您应该这样，所以它留给了编译器.我很惊讶，现代的编译器至少不会向您发出警告.

Basically what gets returned is dumb luck. You get what happens to be in the CPU register when it comes back. If, for example, the returned value would be in AX, and the char* happens to be in AX, you lucked out. I believe this is an undefined behavior; i.e. the C language specifications don't tell what you should so, so it is left to the compiler. I'm surprised a modern compiler wouldn't at least throw a warning at you.

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