分配OPTARG在C的int [英] Assigning optarg to an int in C

问题描述

我想一个OPTARG值赋给一个int，但是编译器给了我以下警告：

 警告：赋值时将指针整数，未作施放[默认启用]
 

我试图铸造OPTARG为int的赋值之前

  N =（INT）OPTARG;
 

但仍得到一个警告：

 警告：从指针转换为大小不同的整数[-Wpointer对INT-投]
 

我不知道需要之前，我可以简单地OPTARG分配到一个整数，然后打印（现在）做什么。

  INT主（INT ARGC，CHAR *的argv []）
{
  焦炭℃;
  INT N;  而（（c = getopt的（ARGC，ARGV，M））！= -1）{
    开关（三）{
    案件的m：
      N = OPTARG;
      打破;
    }
  }  的printf（％d个\\ N，N）;  返回0;
}
 

解决方案

选项字符串的总是的字符串。

如果你想要一个整数，你需要使用一个转换功能，像的与atoi（3）

所以，你应该至少code

  N =的atoi（OPTARG）;
 

当心， OPTARG 可能是 NULL ，当然可能是一个非数字。您可以使用与strtol（3）可以设置结束字符你会检查。

因此​​，一个更严重的办法是

 情况下，M：
  {
     字符* ENDP = NULL;
     长L = -1;
     如果（OPTARG ||（（L =与strtol（OPTARG，0，＆安培;！ENDP）），（ENDP和放大器;＆放大器; * ENDP）））
       {fprintf中（标准错误，无效-m选项％S  - 期待一个数字\\ n
                 OPTARG OPTARG：）;
         出口（EXIT_FAILURE）;
       };
      //你可以开左增加更多的检查，这里...
      N =（INT）升;
     打破;
  }
  N = OPTARG;
  打破;
 

^{注意分配到→如前pression和的在如果测试里面逗号操作。}

顺便说一句，GNU库也有 argp 功能（和还 getopt_long - 但 argp 功能更强大的），您可能会发现更方便。几个框架（尤其是基于GTK和Qt）也有程序参数传递的功能。

如果你正在做一个严肃的节目，请让它接受 - 帮助选项，如果可能的话 - 版本之一。这真的很方便，我恨它不接受他们的一些程序。看看 GNU标准说。

I am trying to assign an optarg value to an int, but the compiler gives me the following warning:

warning: assignment makes integer from pointer without a cast [enabled by default]

I have tried casting optarg as int before the assignment

n = (int) optarg;

but still get a warning:

warning: cast from pointer to integer of different size [-Wpointer-to-int-cast]

I am not sure what needs to be done before I can simply assign the optarg to an integer, then print it (for now).

int main (int argc, char *argv[])
{
  char c;
  int n;

  while ((c = getopt(argc, argv, "m:")) != -1) {
    switch (c) {
    case 'm':
      n = optarg;
      break;
    }
  }

  printf("%d\n", n);

  return 0;
}

解决方案

The option string is always a string.

If you want an integer, you need to use a conversion function, like atoi(3)

So you should at least code

n = atoi(optarg);

Beware, optarg could be NULL and could certainly be a non-number. You might use strtol(3) which may set the ending character which you would check.

So a more serious approach could be

case 'm':
  {
     char* endp = NULL;
     long l = -1;
     if (!optarg ||  ((l=strtol(optarg, 0, &endp)),(endp && *endp)))
       { fprintf(stderr, "invalid m option %s - expecting a number\n", 
                 optarg?optarg:"");
         exit(EXIT_FAILURE);
       };
      // you could add more checks on l here...
      n = (int) l;
     break;
  }
  n = optarg;
  break;

^{Notice the assignment to l as expression and the comma operator inside the if test.}

BTW, GNU Libc also have argp functions (and also getopt_long - but argp functions are more powerful), which you may find more convenient. Several frameworks (notably Gtk and Qt) have also program argument passing functionalities.

If you are doing a serious program please make it accept the --help option, and if possible the --version one. It is really convenient, and I hate the few programs which don't accept them. See what GNU standards say.

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