在一行中在 main() 中初始化/设置 char *argv[] [英] Initialize/set char *argv[] inside main() in one line
问题描述
我想在 main() 中初始化/设置 char *argv[] 以便我可以使用 argv[1], argv[2]... 稍后在我的程序中.
I want to initialize/set char *argv[] inside the main() so that I can use argv[1], argv[2]... later in my program.
到目前为止，我知道如何通过两种方式做到这一点:
Up to now, I know how to do this in two ways:
对于
int main()，使用一行:
int main()
{
char *argv[] = {"programName", "para1", "para2", "para3", NULL};
}
注意，最后使用NULL是因为argv数组中的指针指向C字符串，根据定义，NULL终止.
Note that, using NULL in the end is because the pointers in the argv array point to C strings, which are by definition NULL terminated.
对于int main(int argc, char* argv[])，我必须使用多行作为:
For int main(int argc, char* argv[]), I have to use multiple lines as:
int main(int argc,char* argv[])
{
argv[0] = "programName";
argv[1] = "para1";
argv[2] = "para2";
argv[3] = "para3";
}
我的问题是如何将这两种方法结合在一起，即仅使用一行来为 int main(int argc, char* argv[]) 初始化它?特别是，我希望能够这样做(目前这将是错误的):
My question is that how can I combine these two methods together, i.e. use only one line to initialize it for int main(int argc, char* argv[])? Particularly, I want to be able to do like this (this will be wrong currently):
int main(int argc, char* argv[])
{
argv = {"programName", "para1", "para2", "para3", NULL};
}
我怎样才能做到这一点?
How can I be able to do this?
编辑:我知道argv[] 可以在Debugging Command Arguments 中设置.我想在 main() 中编辑它们的原因是我不想每次都为一个新的测试用例(不同的 Debugging Command Arguments代码>argv[] 设置).
Edit: I know argv[] can be set in Debugging Command Arguments. The reason that I want to edit them in main() is that I don't want to bother to use Debugging Command Arguments every time for a new test case (different argv[] setting).
推荐答案
最安全的方法可能是不要写入 argv 引用的内存，(可能不是你想的那样结构化)，而是有另一个批量:
The safest way is probably don't write into argv referred memory, (that may not be structured as you think), but having another bulk:
int main(int argc, const char** argv)
{
const char* n_argv[] = { "param0", "param1", "param2" };
argv = n_argv;
...
}
这会将 argv 从它的原始内存(由调用者拥有，保留在那里)转移到另一个将在 main() 的生命周期中存在的内存.
This will distract argv from it original memory (owned by the caller, that remains there) to another that will exist for the life of main().
请注意，需要 const char* 以避免不推荐使用的*string分配给 char**"消息.
Note that const char* is required to avoid the deprecated "*string assigned to char**" message.
注意:此代码已在 Linux 上使用 GCC 4.8.1 编译，结果如下:
NOTE: this code had been compiled with GCC 4.8.1 on Linux giving the following results:
make all
Building file: ../main.cpp
Invoking: GCC C++ Compiler
g++ -O0 -g3 -pedantic -Wall -c -std=c++11 -o "main.o" "../main.cpp"
Finished building: ../main.cpp
Building target: e0
Invoking: GCC C++ Linker
g++ -o "e0" ./main.o
Finished building target: e0
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