Setting values in a 2D array in C++

Question

I've tried to create a 2D array where each row is going to hold a texture/image values (RGBA) (where each column of the texture will be placed after each other in a row). The base for my texture will be all white, therefore I set all values to 1. However, when I print out the value it is 0 and not 1. Why does data[layer][x*m_Size+y+a] = 1; not set the value of that position to 1?

data = new int*[m_Size];
for(unsigned i = 0; i<m_Size; ++i){
    data[i] = new int[m_Size*m_Size*4];
}
for(unsigned layer=0; layer<m_NumLayers; layer++){  
    for (unsigned x = 0; x < m_Size; x++){
        for (unsigned y = 0; y < m_Size*4; y+=4){
                data[layer][x*m_Size+y+r] = 1;  
                data[layer][x*m_Size+y+g] = 1;  
                data[layer][x*m_Size+y+b] = 1;  
                data[layer][x*m_Size+y+a] = 1;  
                printf("in data: %f \n,",data[layer][x*m_Size+y+a]);
        }

    }

}

If m_NumLayers is more than m_Size, then your loop will overflow data. — eerorika
– eerorika, Commented Jan 7, 2016 at 12:14
Your printf call is wrong. You want printf("in data: %d \n,",data[layer][x*m_Size+y+a]); instead of printf("in data: %f \n,",data[layer][x*m_Size+y+a]); — ThorngardSO
– ThorngardSO, Commented Jan 7, 2016 at 12:16

ThorngardSO · Accepted Answer · 2016-01-07 12:22:53Z

1

Your printf call is wrong. You want printf("in data: %d \n,",data[layer][x*m_Size+y+a]); instead of printf("in data: %f \n,",data[layer][x*m_Size+y+a]);, because you want to print an integer argument.

answered Jan 7, 2016 at 12:22

ThorngardSO

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Comments

xinaiz · Accepted Answer · 2016-01-07 12:22:42Z

0

You should use std::cout instead of printf. You see, you messed type modifier in printf - %f is for float, %d is for int.

Use cout instead:

#include <iostream>
...
std::cout << "in data: " << data[layer][x*m_Size+y+a] << std::endl;

answered Jan 7, 2016 at 12:22

xinaiz

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