如何让GCC警告无与伦比的数量的参数使得函数调用时？ [英] How to let GCC warn unmatched number of arguments when making function calls?

问题描述

我刚刚调试一个C程序很长一段时间，才发现我做一个函数调用时错过了一个说法，所以垃圾，而不是填补缺少的参数。这样愚蠢的错误是很无奈，但我想编制者应能够检测到这一点。 （C甚至不支持默认参数，即使在C ++中，默认参数需要显式声明）

更新：原型被认为是错误的，太...

那么，是有无与伦比的警告函数调用的参数数量的GCC标志？我一直有 -Wall 和 -pedantic 上;这是相当令人惊讶，这样一个明显的错误未被发现。 （其实我想有一些原因，GCC不报告，但我不能在这个时候想到的任何。）

尴尬code例如：

 静态无效DFS（）;    INT主（INT ARGC，为const char * argv的[]）{
         DFS（1）;
    }    静态无效
    DFS（INT依然存在，去年INT）{
        // DFS
    }
 

我只是做了另一项发现是，如果原型包含参数，编译器将报告;但原型碰巧包含任何参数，那么编译器只是下滑。

解决方案

函数调用参数不匹配的号码是强制性的诊断，所有的编译器都必须提供没有任何特殊的设置。它是由标准规定的。

C99Standard 6.5.2.2函数调用：结果
限制

  

如果它表示所调用的函数的前pression有一个类型，其中包括一个原型，该
  参数的数目应的参数的数量一致。每个参数应
  有一种类型，使得它的值可以被分配给一个对象与不合格版本
  的其对应的参数的类型。

---

 静态无效DFS（）;
 

告诉编译器 DFS 是静态函数返回一个无效，可以采取的参数数目不详。此外，您提供哪些需要两个参数和放大器功能的定义;调用相同。正如你看到的，没有合同的断裂。问题是函数的声明不正确。如果要声明一个函数，它不带任何参数，你必须使用：

 静态无效DFS（无效）;
 

一旦你这样做的 编译器将为您提供一个诊断 。

I just debugged a C program for a long time, only to find that I missed an argument when making a function call, so junk instead filled the missing argument. Stupid mistakes like this are really frustrating, but I suppose compilers should be able to detect this. (C doesn't even support default arguments; even in C++, default arguments need to be explicitly declared.)

Update: The prototype was found to be wrong, too...

So, is there a GCC flag for warning unmatched function call argument number? I always have -Wall and -pedantic on; it's quite surprising that such an obvious error goes undetected. (Actually I suppose there is some reason that GCC does not report, but I can't think of any at this time.)

Embarrassing code example:

    static void dfs();

    int main(int argc, const char *argv[]) {
         dfs(1);
    }

    static void
    dfs(int remain, int last) {
        // dfs
    }

Another discovery I just made is that if the prototype contains argument, the compiler will report; but the prototype happened to contains no arguments, then the compiler just slipped.

解决方案

Unmatched number of function call arguments is a mandatory diagnostic that all compilers will and must provide without any special setting. It is mandated by the standard.

C99Standard 6.5.2.2 Function calls:
Constraints

If the expression that denotes the called function has a type that includes a prototype, the number of arguments shall agree with the number of parameters. Each argument shall have a type such that its value may be assigned to an object with the unqualified version of the type of its corresponding parameter.

---

 static void dfs();

Tells the compiler dfs is a static function which returns a void and can take unspecified number of arguments. Further you provide a definition for the function which takes 2 arguments & call the same. As you see there is no breaking of contract. The problem is the declaration of the function is incorrect. If you want to declare a function which takes no arguments you must use:

 static void dfs(void);

Once you do that the compiler will provide you a diagnostic.

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