Visual C ++错误的切入点 [英] Visual C++ wrong entry point

问题描述

我遇到一个问题，启动使用

编译的应用程序Visual C ++ 6 SP5不执行函数main（int argc，char

** argv）并退出立即使用代码1（0x1）。


将main作为项目中的入口点 - >设置 - >

链接 - >输出 - >入口点符号，使程序调用main

但我的argc和argv参数都等于0.


有人有想法吗？

I experience a problem where starting an application compiled with
visual C++ 6 SP5 does not execute the function main(int argc, char
**argv) and exit immediately with code 1 (0x1).

Putting main as an entry point in the section Project -> settings ->
link -> output -> entry-point symbol, makes the program calling main
but my argc and argv arguments are both equal to 0.

Has anyone an idea ?

推荐答案

" Laurent Schall" < SC ****** @ yahoo.fr>在消息中写道

新闻：a9 ************************** @ posting.google.c om ...

"Laurent Schall" <sc******@yahoo.fr> wrote in message
news:a9**************************@posting.google.c om...

我遇到一个问题，启动使用Visual C ++ 6 SP5编译的应用程序不执行函数main（int argc，char
** argv）并立即退出代码1（0x1）。

将main作为入口点放在Project - >中。设置 - >
链接 - >输出 - >入口点符号，使程序调用main
但我的argc和argv参数都等于0.

有人有想法吗？

I experience a problem where starting an application compiled with
visual C++ 6 SP5 does not execute the function main(int argc, char
**argv) and exit immediately with code 1 (0x1).

Putting main as an entry point in the section Project -> settings ->
link -> output -> entry-point symbol, makes the program calling main
but my argc and argv arguments are both equal to 0.

Has anyone an idea ?

也许你的程序包含构造函数失败的全局对象。这个听起来有点合理吗？


设置项目设置下的入口点听起来真的很糟糕

想法，你只是弄乱你不理解的东西。


john

Maybe your program contains global objects whose constructors fail. Does
this sound plausible?

Setting the entry point under project settings sounds like a really bad
idea, you were just messing with something you didn''t understand.

john

* sc ****** @ yahoo.fr （Laurent Schall）schriebt：

* sc******@yahoo.fr (Laurent Schall) schriebt:

我遇到启动应用程序的问题使用Visual C ++编译的SP5不执行函数main（int argc，char
** argv）并立即退出代码1（0x1）。


正如John Harrison回应的那样，也许你的程序包含全局对象

，其构造函数失败。


另一种可能性可能是你使用了那个编译器的非标准的

功能而不是''main''''WinMain''或者其他一些（实际上是

执行几种可能性;在这种情况下，更改项目设置

或适当修改编译器/链接器参数;请参阅您的文档。


将main作为项目中的入口点 - >设置 - >
链接 - >输出 - >入口点符号，使程序调用main
但我的argc和argv参数都等于0.

有人有想法吗？

I experience a problem where starting an application compiled with
visual C++ 6 SP5 does not execute the function main(int argc, char
**argv) and exit immediately with code 1 (0x1).
As John Harrison responded, perhaps your program contains global objects
whose constructors fail.

Another possibility might be that you have used that compiler''s non-standard
feature where instead of ''main'' a ''WinMain'' or some such (there are actually
several possibilities) is executed; in that case change the project settings
or modify the compiler/linker arguments appropriately; see your documentation.

Putting main as an entry point in the section Project -> settings ->
link -> output -> entry-point symbol, makes the program calling main
but my argc and argv arguments are both equal to 0.

Has anyone an idea?

是的，这是该组中的偏离主题，但是为了让你回到

track：你指定给链接器的入口点，几乎无论

哪个C ++实现，不是''main''函数，但是在C ++实现的运行时库中有些函数是b $ b，而后者又调用''main''。 />

Yes, that is off-topic in this group, but in order to get you back on
track: the entry point you specify to the linker, almost regardless of
which C++ implementation, is _not_ the ''main'' function, but some function
in that C++ implementation''s runtime library that in turn calls ''main''.

Laurent Schall写道：

Laurent Schall wrote:

我遇到了启动使用Visual C ++编译的应用程序的问题6 SP5不执行函数main（int argc，char
** argv）并立即退出代码1（0x1）。

将main作为入口点放在Project - >中;设置 - >
链接 - >输出 - >入口点符号，使程序调用main
但我的argc和argv参数都等于0.

有人有想法吗？

I experience a problem where starting an application compiled with
visual C++ 6 SP5 does not execute the function main(int argc, char
**argv) and exit immediately with code 1 (0x1).

Putting main as an entry point in the section Project -> settings ->
link -> output -> entry-point symbol, makes the program calling main
but my argc and argv arguments are both equal to 0.

Has anyone an idea ?

Main定义为：


int main（int argc，char * argv []）


我不喜欢知道这是否能解决你的问题，但这是一个开始。

Main is defined as:

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

I don''t know if that will solve your problem or not, but it is a start.

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