malloc 2D array

hi all

I am using the following code, which is printing a value of 0.0 instead of 212.0. What am I doing wrong?

Code:

---


double **u;
int N = 52;
int i;


u = malloc (N * sizeof(double *));;
for (i = 0; i < N; i++)
    u[i] = malloc(N * sizeof(double));

u[10][10] = 212.0;

printf("U[10][10] = %4.3f", u[10][10]);

---

It looks right to me. I think your problem must be somewhere else.

You need to provide us with additional details? Is all that code part of main() or is it part of different functions? Is **u a global variable?

I have copied your code and I executed like the followings.It worked correctly for me.

Code:

---

#include<stdio.h>
#include<malloc.h>
main()
{
double **u;
int N = 52;
int i;


u = malloc (N * sizeof(double *));;
for (i = 0; i < N; i++)
u[i] = malloc(N * sizeof(double));

u[10][10] = 212.0;

printf("U[10][10] = %4.3f", u[10][10]);
}

---

Output

U[10][10] = 212.000

I too tried your code. Thats work for me.

Remember to use free() on what you malloc().

deleted, was wrong

Quote:

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Originally Posted by Epy

%f is for float, %lf is for double, you're using a double.

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DerekC used printf though, not scanf, so %f for double is correct.

Quote:

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Originally Posted by laserlight

DerekC used printf though, not scanf, so %f for double is correct.

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Damnit. Thanks, maybe I need to lay off the forums until I wake up.

I actually tried exactly what you in that main function, all by itself,and didnt get a correct answer. Let me double check one last time, and I will be back in a sec.

Ah. I tracked it down. I had to include stdio.h and stdlib.h to supress the errors about casting the malloc as a (double**) and get the correct return. Thanks for the help..I didnt think I needed that