指向字符的指针 [英] Pointer to Pointer to character
问题描述
一个简单的程序...
#include< iostream>
使用命名空间std;
int main（int argc，char ** argv）{
cout<< argv<< endl;
cout<< ; ** argv<< endl;
cout<< argv [0] [0]<< endl;
cout<<& argv [ 0]<< endl;
cout<<&（argv [0] [0]）<< endl; //为什么不和& argv [0]相同？
}
为什么& argv [0]& &（argv [0] [0]）差异？
a simple program...
#include <iostream>
using namespace std;
int main(int argc, char** argv) {
cout<<argv<<endl;
cout<<**argv<<endl;
cout<<argv[0][0]<<endl;
cout<<&argv[0]<<endl;
cout<<&(argv[0][0])<<endl; // why not the same as &argv[0]?
}
why &argv[0] & &(argv[0][0]) difference?
推荐答案
Howachen发布：
Howachen posted:
#include< iostream>
#include <iostream>
您需要包含ostream如果你想使用endl。
You need to include "ostream" if you want to use "endl".
using namespace std;
int main（int argc，char ** argv）{
cout<< argv<< endl;
using namespace std;
int main(int argc, char** argv) {
cout<<argv<<endl;
我不认为ostream :: operator<<有一个过载需要一个char **。
I don''t think ostream::operator<< has an overload which takes a char**.
cout<< ** argv<< endl;
cout<<**argv<<endl;
这将打印一个字符。
This will print a single char.
cout<< argv [0 ] [0]<< ENDL;
cout<<argv[0][0]<<endl;
就像这样。 （因为它相当于：
*（*（argv + 0）+ 0）
等于：
** argv
As will this. (Because it''s equivalent to:
*(*(argv + 0) + 0)
which is equal to:
**argv
cout<<& argv [0]<< endl;
cout<<&argv[0]<<endl;
这相当于：
& *（argv + 0）
等于：
argv + 0
等于：
argv
This is equal to:
&*(argv+0)
which is equal to:
argv+0
which is equal to:
argv
cout<<&（argv [0] [0]）<< endl; //为什么不一样as& argv [0]？
cout<<&(argv[0][0])<<endl; // why not the same as &argv[0]?
这相当于：
& *（*（argv + 0 ）+ 0）
等于：
*（argv + 0）
等于：
* argv
-
Frederick Gotham
This is equal to:
&*(*(argv+0)+0)
which is equal to:
*(argv+0)
Which is equal to:
*argv
--
Frederick Gotham
howachen写道：
howachen wrote:
why& argv [0]&&（argv [0] [0不同的e？
why &argv[0] & &(argv[0][0]) difference?
因为argv [0]是char *类型而argv [0] [0]是char类型。所以他们的
地址分别是char **和char *。
-
Phlip
http://www.greencheese.us/ZeekLand < - 不是博客!!!
Because argv[0] is type char * and argv[0][0] is type char. So their
addresses are char** and char* respectively.
--
Phlip
http://www.greencheese.us/ZeekLand <-- NOT a blog!!!
howachen写道：
howachen wrote:
> why& argv [0 ]& &（argv [0] [0]）有什么区别？
>why &argv[0] & &(argv[0][0]) difference?
因为argv [0]是char *类型而argv [0] [0]是char类型。所以他们的
地址分别是char **和char *。
Because argv[0] is type char * and argv[0][0] is type char. So their
addresses are char** and char* respectively.
另外因为argv [0]是指针，所以指针本身的地址在这里是
，而argv [0] [ 0]是一个字符，所以它的地址就在那里。
断言（argv [0] ==& argv [0] [0]）;
-
Phlip
http： //www.greencheese.us/ZeekLand < - 不是博客!!!
Also because argv[0] is a pointer, so the address of the pointer itself is
here, while argv[0][0] is a character, so its address is there.
assert (argv[0] == &argv[0][0]);
--
Phlip
http://www.greencheese.us/ZeekLand <-- NOT a blog!!!
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