如何从 main() 函数外部访问 argv[]？

发布于 2024-09-30 01:45:06 字数 475 浏览 4 评论 0原文

我碰巧有几个函数可以通过 argv[] 数组访问程序的不同参数。目前，这些函数嵌套在 main() 函数中，因为编译器提供了允许此类结构的语言扩展。

我想摆脱嵌套函数，以便在不依赖语言扩展的情况下实现互操作性。

首先，我想到了一个数组指针，一旦程序启动，我将指向 argv[] ，该变量将位于 main() 函数之外并在之前声明功能，以便他们可以使用它。

所以我声明了这样一个指针，如下所示：

char *(*name)[];

它应该是一个指向字符指针数组的指针。但是，当我尝试将其指向 argv[] 时，我收到关于不兼容指针类型的赋值的警告：

name = &argv;

可能是什么问题？您是否想到了另一种从 main() 函数外部访问 argv[] 数组的方法？

原文

I happen to have several functions which access different arguments of the program through the argv[] array. Right now, those functions are nested inside the main() function because of a language extension the compiler provides to allow such structures.

I would like to get rid of the nested functions so that interoperability is possible without depending on a language extension.

First of all I thought of an array pointer which I would point to argv[] once the program starts, this variable would be outside of the main() function and declared before the functions so that it could be used by them.

So I declared such a pointer as follows:

char *(*name)[];

Which should be a pointer to an array of pointers to characters. However, when I try to point it to argv[] I get a warning on an assignment from an incompatible pointer type:

name = &argv;

What could be the problem? Do you think of another way to access the argv[] array from outside the main() function?

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痴梦一场 2024-10-07 01:45:06

char ** name;
...
name = argv;

就可以了:)

你会看到 char *(*name) [] 是一个指向 char 指针数组的指针。而函数参数 argv 的类型为指向 char 的指针，因此 &argv 的类型为指向 char 的指针。为什么？因为当您声明一个函数接受数组时，对于编译器来说，它与接受指针的函数相同。也就是说，

void f(char* a[]);
void f(char** a);
void f(char* a[4]);

是完全相同的等效声明。并不是说数组是指针，而是作为函数参数它是

HTH

char ** name;
...
name = argv;

will do the trick :)

you see char *(*name) [] is a pointer to array of pointers to char. Whereas your function argument argv has type pointer to pointer to char, and therefore &argv has type pointer to pointer to pointer to char. Why? Because when you declare a function to take an array it is the same for the compiler as a function taking a pointer. That is,

void f(char* a[]);
void f(char** a);
void f(char* a[4]);

are absolutely identical equivalent declarations. Not that an array is a pointer, but as a function argument it is

HTH

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小…红帽 2024-10-07 01:45:06

这应该有效，

char **global_argv;


int f(){
 printf("%s\n", global_argv[0]); 
}

int main(int argc, char *argv[]){
  global_argv = argv;
 f(); 
}

This should work,

char **global_argv;


int f(){
 printf("%s\n", global_argv[0]); 
}

int main(int argc, char *argv[]){
  global_argv = argv;
 f(); 
}

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ぇ气 2024-10-07 01:45:06

#include <stdio.h>

int foo(int pArgc, char **pArgv);

int foo(int pArgc, char **pArgv) {
    int argIdx;

    /* do stuff with pArgv[] elements, e.g. */      
    for (argIdx = 0; argIdx < pArgc; argIdx++)
        fprintf(stderr, "%s\n", pArgv[argIdx]);

    return 0;
}

int main(int argc, char **argv) {
    foo(argc, argv);
}

#include <stdio.h>

int foo(int pArgc, char **pArgv);

int foo(int pArgc, char **pArgv) {
    int argIdx;

    /* do stuff with pArgv[] elements, e.g. */      
    for (argIdx = 0; argIdx < pArgc; argIdx++)
        fprintf(stderr, "%s\n", pArgv[argIdx]);

    return 0;
}

int main(int argc, char **argv) {
    foo(argc, argv);
}

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