使用char *所需的字符串 [英] Using string where a char* required
问题描述
我正在使用C API，它通常需要char *或char缓冲区。如果一个
C函数返回一个char *我不能使用字符串？或者我可以吗？我意识到我可以使用c_str（）传递一个char *，但是接收一个char缓冲区到一个
字符串呢？
我问的原因是否则我要猜测要创建什么尺寸的缓冲区？
然后将缓冲区复制到字符串。看起来并不理想。
" Angus"写道：
>我正在使用C API，它通常需要char *或char缓冲区。如果
C函数返回一个char *我不能使用字符串？或者我可以吗？我意识到我可以使用c_str（）传递一个char *，但是接收一个char缓冲区到一个
字符串呢？
我问的原因是否则我要猜测要创建什么尺寸的缓冲区？
然后将缓冲区复制到字符串。看起来不太理想。
std :: string的构造函数之一实际上会接受一个char *作为
参数。
因此：
char * foo {/ * stuff * /}
std :: string s;
s = foo（）;
或者至少类似的东西。
" osmium" < r1 ******** @ comcast.netwrote in message
news：5h ************* @ mid.individual.net ...
" Angus"写道：
我正在使用C API，它通常需要char *或char缓冲区。
如果
a
C函数返回一个char *我不能用字符串？或者我可以吗？我意识到我
可以
通过char *使用c_str（）但是接收一个char缓冲区到
字符串怎么样？
原因我问的是否则我要猜测缓冲区的大小创建？
然后将缓冲区复制到字符串。看起来不太理想。
std :: string的一个构造函数实际上会接受一个char *作为
参数。
因此：
char * foo {/ * stuff * /}
std :: string s;
s = foo（）;
或者至少类似的东西。
For例如在Windows中有一个GetUserName函数，如下所示：
BOOL WINAPI GetUserName（LPWSTR lpBuffer，LPDWORD nSize）
如果你这样做：
std :: string str;
DWORD bufsize = 256;
GetUserName（str，& bufsize）;
你得到编译错误：
错误C2664：''GetUserNameA''：无法从''类转换参数1
std :: basic_string< char，struct std :: char_traits< char>，class
std :: allocator< char''to''char *''
没有可用的用户定义转换运算符可以执行此
con版本，或操作员不能被调用
我想我可以演员？但想知道最好的解决方法是什么。
Angus写道：
" osmium" ; < r1 ******** @ comcast.netwrote in message
news：5h ************* @ mid.individual.net ...
>" Angus"写道：
>>我正在使用一个C API，它通常需要一个char *或char
缓冲区。如果一个
C函数返回一个char *我不能使用字符串？或者我可以吗？我知道我可以使用c_str（）传递一个char *但是接收一个字符串的
char缓冲区怎么样？
我问的原因是否则我必须猜测是什么大小缓冲区来创建？然后将缓冲区复制到字符串。看起来不太理想。
std :: string的一个构造函数实际上会接受一个
char *作为参数。
因此：
char * foo {/ * stuff * /}
std :: string s;
s = foo（）;
或者在至少类似的东西。
例如在Windows中有一个像这样的GetUserName函数：
BOOL WINAPI GetUserName（LPWSTR lpBuffer，LPDWORD nSize）
如果你这样做：
std :: string str;
DWORD bufsize = 256;
GetUserName（str，& bufsize）;
你得到编译错误：
错误C2664：''GetUserNameA''：无法从''类转换参数1
std :: basic_string< char，struct std :: char_traits< char>，class
std :: allocator< char''to''char *''
没有可用的用户定义转换运算符，可以执行此转换，或者运算符不能呼叫ed
我想我可以演员？但想知道最好的解决方法是什么。
最好的方法是分配TCHAR的临时（本地）数组（或者什么类型是LPWSTR指向的b / b
）然后_after_你调用
函数来填充该数组，从中创建一个字符串。
V
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I am working with a C API which often requires a char* or char buffer. If a
C function returns a char* I can''t use string? Or can I? I realise I can
pass a char* using c_str() but what about receiving a char buffer into a
string?
Reason I ask is otherwise I have to guess at what size buffer to create?
And then copy buffer to a string. Doesn''t seem ideal.
"Angus" writes:
>I am working with a C API which often requires a char* or char buffer. If
a
C function returns a char* I can''t use string? Or can I? I realise I can
pass a char* using c_str() but what about receiving a char buffer into a
string?
Reason I ask is otherwise I have to guess at what size buffer to create?
And then copy buffer to a string. Doesn''t seem ideal.One of the constructors for std::string will, in effect, accept a char* as
an parameter.
Thus:
char* foo { /*stuff*/ }
std::string s;
s= foo();
Or at least something similar to that.
"osmium" <r1********@comcast.netwrote in message
news:5h*************@mid.individual.net..."Angus" writes:
I am working with a C API which often requires a char* or char buffer.
If
a
C function returns a char* I can''t use string? Or can I? I realise I
can
pass a char* using c_str() but what about receiving a char buffer into a
string?
Reason I ask is otherwise I have to guess at what size buffer to create?
And then copy buffer to a string. Doesn''t seem ideal.
One of the constructors for std::string will, in effect, accept a char* as
an parameter.
Thus:
char* foo { /*stuff*/ }
std::string s;
s= foo();
Or at least something similar to that.
For example in Windows there is a GetUserName function like this:
BOOL WINAPI GetUserName ( LPWSTR lpBuffer, LPDWORD nSize )
if you do this:
std::string str;
DWORD bufsize = 256;
GetUserName(str, &bufsize);
you get compile error:
error C2664: ''GetUserNameA'' : cannot convert parameter 1 from ''class
std::basic_string<char,struct std::char_traits<char>,class
std::allocator<char'' to ''char *''
No user-defined-conversion operator available that can perform this
conversion, or the operator cannot be called
I suppose I could cast? But wondering what best way to tackle is.
Angus wrote:"osmium" <r1********@comcast.netwrote in message
news:5h*************@mid.individual.net...>"Angus" writes:
>>I am working with a C API which often requires a char* or char
buffer. If a
C function returns a char* I can''t use string? Or can I? I
realise I can pass a char* using c_str() but what about receiving a
char buffer into a string?
Reason I ask is otherwise I have to guess at what size buffer to
create? And then copy buffer to a string. Doesn''t seem ideal.
One of the constructors for std::string will, in effect, accept a
char* as an parameter.
Thus:
char* foo { /*stuff*/ }
std::string s;
s= foo();
Or at least something similar to that.
For example in Windows there is a GetUserName function like this:
BOOL WINAPI GetUserName ( LPWSTR lpBuffer, LPDWORD nSize )
if you do this:
std::string str;
DWORD bufsize = 256;
GetUserName(str, &bufsize);
you get compile error:
error C2664: ''GetUserNameA'' : cannot convert parameter 1 from ''class
std::basic_string<char,struct std::char_traits<char>,class
std::allocator<char'' to ''char *''
No user-defined-conversion operator available that can perform
this conversion, or the operator cannot be called
I suppose I could cast? But wondering what best way to tackle is.The best way is to allocate a temporary (local) array of TCHAR (or
whatever type is LPWSTR points to) and then _after_ you call the
function that would fill in that array, create a string from it.
V
--
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