如果const char *在函数调用中转换为char *，则可能出现任何问题 [英] Any possible issues if const char * is converted to char * in a function call

问题描述

在fun1（）中我得到一个字符串，我需要将这个字符串数据传递给函数fun2（），它接受参数char *



function2 prototype is如下所示



void fun2（char * p）;

我从fun1调用fun2，如下所示



void fun1（）

{

fun2（（char *）str.c_str（））;

}

从const char *到char *的转换是否有任何可能造成任何潜在问题

In fun1() I am getting a string and I need to pass this string data to a function fun2() which takes argument char *

function2 prototype is as below

void fun2(char *p);
I am calling fun2 from fun1 as below

void fun1()
{
fun2((char *)str.c_str());
}
Is there any chance from the conversion from const char * to char * does make any potential issues

推荐答案

是。

const char *无法转换为char *因为它会打开更改常量值的可能性：在这种情况下可能会改变（比方说）用户提示符来自hello user！ HeLlO UsEr！并且在程序中途进行固定提示更改。

如果需要将const char *传递给带char *的函数，则需要分配一些char内存，并复制常量字符串进入它 - 然后你可以将非const副本传递给你的函数。

Yes.
A const char* can't be converted to a char* because it would open up the possiblity of changing the value of constants: in this case potentially changing (say) a user prompt from "hello user!" to "HeLlO UsEr!" and making the fixed prompt change half way through the program.
If you need to pass a const char* to a function taking a char*, you need to allocate some char memory, and copy the constant string into it - then you can pass the non-const copy to your function.

你可以通过简单地获取第一个字符的地址从C ++字符串中获取可修改的C字符串。使用

You can get a modifiable C string from a C++ string by simply taking the address of the first character. Use

void fun1()
{
    fun2(&str[0]);
}

（唯一的问题是如果fun2碰巧修改了其C字符串参数的终止空字符，则为null在这种情况下附加的C ++字符串是不可修改的）

(the only catch is if fun2 happens to modify the terminating null character of its C string argument, the null appended by the C++ string in this situation is not modifiable)

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