using a string array in a function C++ - c++

Question

this question should be easy and straight forward, but after searching online, I couldn't find an answer. might because the question is just too simple.
following code is from cplusplus.com. it's a function of making a string lowercase. I was intended to do something similar.
/* tolower example */
#include <stdio.h>
#include <ctype.h>
int main ()
{
int i=0;
char str[]="Test String.\n";
char c;
while (str[i])
{
c=str[i];
putchar (tolower(c));
i++;
}
return 0;
}
and what I made is this:
void search(string A[], string B[], int k)
{
int temp;
for(int j = 0; j <= 4; j++)
{
for(int i = 0; i <= k; i++)
{
string str (A[i]);
int h = 0;
char lstr[] = B[j];
char c;
while (lstr[h])
{
c = lstr[h];
putchar (tolower(c));
h++;
}
string key (B[j]);
.....
this part of the code is in a for loop. B[j] is a string array.
Visual Studio informed me that char lstr[] = B[j]; part is not right, the error message is:
Error: initialization with '{...}' expected for aggregate object.
I think the problem is that I didn't use the correct syntax of using a string array in a function. something should be done for B[j], in order to make it a char array. I couldn't figure it out.
is that something about pointer? sorry I haven't learn pointer yet.
does my question make sense for you? any help is greatly appreciated!!

Answer 1

If you're looking to make the letters in the string lowercase it's more readable to just work with strings all the way and use std::transform. For example,
// make sure to #include <algorithm>
// at the top
string lstr = B[j];
std::transform(lstr.begin(), lstr.end(), lstr.begin(), ::tolower);
This is much more natural and c++ idiomatic than working with char * directly and less error-prone.

Answer 2

You're trying to assign a char to char[]. You can get the effect you want with the following code:
....
int h = 0;
char* lstr = &B[j]; // point lstr to the address of j'th element of B.
char c;
while (lstr[h])
{
c = lstr[h];
putchar (tolower(c));
h++;
}
.....
What this does is that lstr is now a pointer that points to the j'th character in B. Arrays are essentially pointers. When you do B[j], it's equivalent to writing char ch = *(B + j);, where B points to the address of the first character in the array of characters (otherwise known as string).
EDIT
After your edit, it now seems that you're trying to assign a std::string to a char. Here is the corrected solution.
....
int h = 0;
string& lstr = B[j]; // grab a reference to the j'th string in B.
char c;
while (lstr[h])
{
c = lstr[h];
putchar (tolower(c));
h++;
}
.....
Here, lstr is essentially a reference to the j'th string in B and you can use it as a regular string just like how you're using string str(A[i]);, which makes a copy of the i'th string in A.

Answer 3

You're confusing character arrays and string objects here. A character array is an array of bytes of set size which is null terminated, while a string is an object which expands/contracts as is necessary and doesn't require the null terminator. You're attempting to assign a string object to a character array, which is unsupported. If you're working with string objects, and want to retrieve their equivalent character array, utilize the c_str() function:
const char* lstr = B[j].c_str()
Also, utilizing an array name of B and an index of j is hilarious.