传递参数1会从指针目标类型中丢弃限定词 [英] Passing Argument 1 discards qualifiers from pointer target type

问题描述

我的主要功能如下:

int main(int argc, char const *argv[])
{
    huffenc(argv[1]);
    return 0;
}

编译器返回警告:

huffenc.c:76: warning: passing argument 1 of ‘huffenc’ discards qualifiers from pointer target type

huffenc作为参考，huffenc接受一个char*输入，然后通过./huffenc senselessness

For reference, huffenc takes a char* input, and the function is executed, with the sample input "senselessness" via ./huffenc senselessness

此警告意味着什么?

推荐答案

这意味着您正在将const参数传递给带有非const参数的函数，由于明显的原因，该函数可能很糟糕.

It means that you're passing a const argument to a function which takes a non-const argument, which is potentially bad for obvious reasons.

huffenc可能不需要非const参数，因此应使用const char*.但是，您对main的定义是非标准的.

huffenc probably doesn't need a non-const argument, so it should take a const char*. However, your definition of main is non-standard.

C99标准第5.1.2.2.1节(程序启动)规定:

The C99 standard Section 5.1.2.2.1 (Program startup) states:

在程序启动时调用的函数称为main.实现声明为否 此功能的原型.它应使用返回类型int定义，且不包含 参数:

The function called at program startup is named main. The implementation declares no prototype for this function. It shall be defined with a return type of int and with no parameters:

int main(void) { /* ... */ }

或带有两个参数(此处称为argc和argv，尽管任何名称都可以是 使用，因为它们对于声明它们的函数是局部的):

or with two parameters (referred to here as argc and argv, though any names may be used, as they are local to the function in which they are declared):

int main(int argc, char *argv[]) { /* ... */ }

或等效形式； 9)或以其他实现方式定义的方式.

or equivalent;9) or in some other implementation-defined manner.

接着说...

...参数argc和argv以及argv数组指向的字符串 可由程序修改，并在程序之间保留其最后存储的值 启动和程序终止.

...The parameters argc and argv and the strings pointed to by the argv array shall be modifiable by the program, and retain their last-stored values between program startup and program termination.

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