为什么这个允许（字符*）至（为const char *）推广？ [英] Why does this allow promotion from (char *) to (const char *)?

问题描述

由于scanf函数有来自微软文档中（为const char *）一>和答案<一个href=\"http://stackoverflow.com/questions/766290/shouldnt-it-be-permissible-to-convert-from-char-to-const-char\">this问题中究竟发生了什么会，当我做了（字符**）推广到相同（为const char **）？

Given that scanf has (const char *) in the documentation from Microsoft and the answer to this question what the heck is going when I do the same for (char **) promotion to (const char **)?

基本上，为什么会发生这种编译？

Basically why does this compile?

#include <stdio.h>
int main(int argc, char **argv)
{   
    char szArray[50];
    int  i = 0;
    strcpy(szArray,"10");
    /* the following code is upcasting the (char *) to (const char *) */
    sscanf(szArray,"%d",&i);
    return 0;  
}

为什么不会在编译？

And why won't this compile?

#include <stdio.h>
void processargs(const char **p)
{ 
}
int main(int argc, char **argv)
{
    processargs(argv);          
    return 0;  
}

似乎都在做同样的事情的指针！

Both seem to be doing the same thing to a pointer!

推荐答案

的char ** - ＆GT;为const char ** 是危险的，因为你可能最终意外地修改基础常量对象。

char** -> const char ** is dangerous, since you might end up accidentally modifying the underlying const object.

写你想要什么正确的方法是：

The correct way to write what you want is:

void processargs(const char * const *p)
{ 
}

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