C ++将char转换为const char *

Question

Basically i just want to loop through a string of characters pull each one out and each one has to be of type const char* so i can pass it to a function. heres a example. Thanks for your help.

    string thestring = "abc123";
    const char* theval;
    string result;

    for(i = 0; i < thestring.length(); i++){
        theval = thestring[i]; //somehow convert this must be type const char*
        result = func(theval);
    }

Answer 1

您可以获取该元素的地址：

theval = &thestring[i];

Answer 2

string sym(1, thestring[i]);
theval = sym.c_str();

It gives a null-terminated const char* for every character.

Answer 3

Usually a const char * is pointing to a full null-terminated string, not a single character, so I question if this is really what you want to do.

If it's really what you want, the answer is easy:

theval = &thestring[i];

If the function is really expecting a string but you want to pass it a string of a single character, a slightly different approach is called for:

char theval[2] = {0};
theval[0] = thestring[i];
result = func(theval);

Answer 4

I'm guessing that the func call is expecting a C-string as it's input. In which case you can do the following:

string theString = "abc123";
char tempCString[2];
string result;

tempCString[1] = '\0';

for( string::iterator it = theString.begin();
     it != theString.end(); ++it )
{
   tempCString[0] = *it;
   result = func( tempCString );
}

This will produce a small C-string (null terminated array of characters) which will be of length 1 for each iteration.

The for loop can be done with an index (as you wrote it) or with the iterators (as I wrote it) and it will have the same result; I prefer the iterator just for consistency with the rest of the STL.

Another problem here to note (although these may just be a result of generalizing the code) is that the result will be overwritten on each iteration.

Answer 5

您可以保留该元素的地址：

theval = &thestring[i];