Thread: malloc() and memory leaks

if you want to make a array with malloc take a look here:

Code:

#include <stdio.h>
#include <stdlib.h>
main(){
       int i,j;
       int **X;
           X = (int**)malloc(5*sizeof(int*));
           for(i=0; i<5; i++){
                    X[i]=(int*)malloc(3*sizeof(int));
           }
   for(i=0; i<5; i++){
            for(j=0; j<3; j++){
                     X[i][j]=1;
            }
   }
printf("This is the array: \n \n");   
   for(i=0; i<5; i++){
            printf("\n");
            for(j=0; j<3; j++){
                     printf(" [%d][%d]: %d\t", i, j, X[i][j]);
            }
   }
printf("\n \n");
//system("pause");
}

i hape i help you...

A simple example:

Here we have a function that allocates memory to an integer pointer and assigns a value to it. However we aren't returning the pointer or doing anything else with it. A memory leak has occurred. This memory is still allocated and floating somewhere in the heap, without a way for us to reference it anymore.

Most modern operating systems will free this when the program ends, but there are some out there that won't, in the later case, if we continuously run the program over and over again it will come to a point where you won't be able to allocate any new memory.

Code:

#include <stdio.h>
#include <stdlib.h>

void doSomething();

int main ()
{
   doSomething();
   return 0;
}

void doSomething() 
{
   int* i = malloc(sizeof *i);
   *i = 15;
}