const char * argv [英] const char* argv

问题描述

为什么不是：


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

as：

int main（int argc，const char * argv []）

我假设你不能编辑字符串argv指向。 。 。对吗？

-JKop

Why isn''t:

int main(int argc, char* argv[]) { /* ... */ }
as:
int main(int argc, const char* argv[])
I assume you can''t edit the strings which argv points to. . . right ?
-JKop

推荐答案

" JKop" < NU ** @ NULL.NULL>写在消息>

"JKop" <NU**@NULL.NULL> wrote in message >

为什么不是：

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

as：

int main（int argc，const char * argv []）

我假设你不能编辑argv指向的字符串至。 。 。对吗？

Why isn''t:

int main(int argc, char* argv[]) { /* ... */ }
as:
int main(int argc, const char* argv[])
I assume you can''t edit the strings which argv points to. . . right ?

编号这是C99的报价 -

"参数argc和argv以及argv指向的字符串数组

应该可以由程序修改，并在程序

启动和程序终止之间保留它们最后存储的值

。


Sharad

No. This is a quote from C99 -
"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."

Sharad

不。这是C99的引用 -
The参数argc和argv以及argv
数组指向的字符串应该可由程序修改，并在程序启动和程序终止之间保留它们最后存储的值。

No. This is a quote from C99 -
"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."

这是一个真正的问题：与C ++有什么关系？我认为C99

是某种...... C或C ++的其他方言。 。 。 ？


-JKop

This is a genuine question: What has that to do with C++? I take it that C99
is some sort of... other dialect of C or C++ . . . ?

-JKop

" JKop" < NU ** @ NULL.NULL>在消息中写道

"JKop" <NU**@NULL.NULL> wrote in message

否。这是来自C99的引用 -
参数argc和argv以及argv指向的字符串数组应该可由程序修改，并在程序启动和程序终止之间保留它们最后存储的值。

这是一个真正的问题：有什么与C ++有关？我认为

No. This is a quote from C99 -
"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."

This is a genuine question: What has that to do with C++? I take it that

C99是某种...... C或C ++的其他方言。 。 。 ？

C99 is some sort of... other dialect of C or C++ . . . ?

Nah ...实际上C ++基于C90。您可以在C ++标准中找到对C子条款

的引用。 C99是C标准的最新版本，我从这个引用中引用了b $ b。 AFAIK C ++继承它的方式与C在这个

方面的方式相同，但不太确定。


Sharad

Nah...C++ is based on C90 actually. You can find references to C subclauses
in the C++ Standard. C99 is the more recent revision of the C Standard, I
made a quote from that. AFAIK C++ inherits it the same way as C does in this
regard, not very sure though.

Sharad

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