C++ convert char to const char*
Solution 1
string sym(1, thestring[i]);
theval = sym.c_str();
It gives a null-terminated const char* for every character.
Solution 2
You can take the address of that element:
theval = &thestring[i];
Solution 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);
Solution 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.
user1054513
Updated on October 04, 2020Comments
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user1054513 over 3 years
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); }
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Luchian Grigore over 12 yearsSince he's already using strings, I'm guessing he's just looking for a pointer to char.
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Mark Ransom over 12 years@LuchianGrigore, it's obvious that the function requires a pointer to char but it's not obvious why. It's trivial to pass a single char as a parameter, so it's far more likely that the function takes in a pointer to a null-terminated string.
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John Dibling over 12 yearsNot your downvoter, but this does not answer the question OP actually asked.
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user1054513 over 12 yearsthis is exactly what i needed doing &thestring[i] gave results of abc123, bc123, c123, 123, 23, 3. This gave me each individual character thanks
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WhatsThePoint over 6 yearsCould you please format your code by highlighting it and hitting Ctrl+K
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Max almost 4 yearsOMG thank you so much. For some reason none of the other answers are working lol