issue to populate an array of strings or array of char* - c++

Question

very basic C++ question. Looks I m really rusted there...
All I want to do is to read an array of X strings from a file and create an array of X vertical strings of the horizontal strings.
IE :
file contains:
azert
qsdfg
wxcvb
poiuy
mlkjh
I want to create a string array containing:
aqwpm
zsxol
edcol
rfvuj
tgbyh
Here is what I tried so far:
[bad code]
const int SIZE = 37;
std::string table_h[SIZE];
std::string table_v[SIZE];
int i = 0;
while (source >> table_h[i]) //,sizeof table_h[i]
{
for (int j = 0; j< SIZE; j++)
{
table_v[j][i] = table_h[i][j];
}
i++;
}
-> works fine for the first line, breaks when i=1. I don't understand what.
I noticed that although table_v[0][0] = 'f'; works fine.
Both table_v[0][36] = 'f'; and table_h[0].at(36); break.
With char * (which was my first idea),
char * table_h[SIZE];
char * table_v[SIZE];
something like
table_v[0][0] = 'f';
immediately breaks.
I suppose I need to allocate memory or initialize something first??
Thx in advance.

Answer 1

You should set the size of strings before using operator [] to access them. Resize of table_h is optional, but you definitely have to resize table_v.
const int SIZE = 37;
std::string table_h[SIZE];
std::string table_v[SIZE];
for (size_t i = 0; i < SIZE; ++i)
{
table_h[i].resize(SIZE);
table_v[i].resize(SIZE);
}
int i = 0;
while (source >> table_h[i])
{
for (int j = 0; j < SIZE; j++)
{
table_v[j][i] = table_h[i][j];
}
i++;
}
See the working example.
In my opinion, if you know the size of a strings, resizing is better than appending. It can save some memory re-allocations, and IMHO it is simply nicer solution.

Answer 2

Indeed the table_v[j] is an empty string.
The string needs to allocate space for the characters. This is not done by the index operators, i.e.
table_v[j][9] = 'a';
assumes enough space is allocated for table_v[j].
You can do append to your string to add to the initially empty string. Append does not take chars though, so instead of using index of table_h[i][j] you can use substr.
std::string to_append = table_j[i].substr(j, 1)
table[j].append(to_append);
This also relieves you of the i counter.

Answer 3

Here is a demonstrative program that shows how it can be done
#include <iostream>
#include <vector>
#include <string>
#include <numeric>
int main()
{
std::vector<std::string> v1 =
{
"azert", "qsdfg", "wxcvb", "poiuy", "mlkjh"
};
for ( const std::string &s : v1 ) std::cout << s << ' ';
std::cout << std::endl;
auto max_size = std::accumulate( v1.begin(), v1.end(),
size_t( 0 ),
[]( size_t acc, const std::string &s )
{
return acc < s.size() ? s.size() : acc;
} );
std::vector<std::string> v2( max_size );
for ( const std::string &s : v1 )
{
for ( std::string::size_type i = 0; i < s.size(); i++ )
{
v2[i].push_back( s[i] );
}
}
for ( const std::string &s : v2 ) std::cout << s << ' ';
std::cout << std::endl;
return 0;
}
The program output is
azert qsdfg wxcvb poiuy mlkjh
aqwpm zsxol edcik rfvuj tgbyh
As for your code than these statements
std::string table_h[SIZE];
std::string table_v[SIZE];
defined two arrays of empty strings. So you may not apply the subscript opertaor to an empty string. You could use for example member function push_back
for (int j = 0; j< SIZE; j++)
{
table_v[j].push_back( table_h[i][j] );
}