debug segmentation fault in c++

Question

I need help debugging the following block of code. This block of code gives me a segmentation fault error only when i attempt to assign an appropriate(char) value to squares[ i ][ j ]. How can I fix this?

for(int i=0;i<sides;i++){
    for(int j=0;i<sides;j++){
        squares[i][j] = '';
    }
}

I declared the array squares[][] with the following code before the nested for loop:

char **squares = new(nothrow) char*[sides-1];
for(int i=0; i < sides-1; i++){
    squares[i] = new char[sides-1];
}

you only created sides-1 arrays, but you try to access "sides" number of them in your loop(change the i<sides to i<sides-1) — Creris
– Creris, Commented Apr 24, 2015 at 12:26

Mekap · Accepted Answer · 2015-04-24 12:47:58Z

2

A much better way to declare an empty string array would be to use String[] Don't forget you're in C++. And in your case, since it looks like you're wanting to do a square array, you could use a std::array if the sizes are known or a std::vector for unknown sizes. Now for the strict answer, use j in your second loop instead of i.

for(int i=0;i<sides - 1;i++){
    for(int j=0;j<sides - 1;j++){
        squares[i][j] = '';
    }
}

I'm strongly encouraging you to try to look at your problem with Strings instead of re-inventing the wheel. (Thanks to NathanOliver & jsantander for the help)

edited Apr 24, 2015 at 12:47

answered Apr 24, 2015 at 12:29

Mekap

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4 Comments

jsantander Over a year ago

Don't forget what @Creis said: the loop limits are wrong too.

Mekap Over a year ago

@jsantander yeah, thanks for the reminder, i'm changing that right now

NathanOliver Over a year ago

It would even be better to use a std::array if the sizes are known or a std::vector for unknown sizes.

Mekap Over a year ago

@NathanOliver Updating my answer with that, thanks for your output nathan

胡昊坤 · Accepted Answer · 2015-04-24 12:52:48Z

0

In your declaration,try to change the i < sides - 1 to i < sides

char **squares = new(nothrow) char*[sides-1];
for(int i=0; i < sides; i++){
    squares[i] = new char[sides-1];
}

answered Apr 24, 2015 at 12:52

胡昊坤

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