Segmentation fault by using pointer to pointer

Asked Oct 09 '19 at 14:54

Active Oct 09 '19 at 14:54

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I have the following program which uses a pointer to pointer to print the different elements of a 2D array U.

#include <stdio.h>

int myfunction(double **U) {
    printf("%f\n", U[0][0]); //correctly prints U[0][0]
    printf("%f\n", U[0][1]); //correctly prints U[0][1]
    printf("%f\n", U[1][0]); //"Segmentation fault (core dumped)"
    return 0;
}

int main(void){    
    double U[3][4] = {{10, 1, 0, 0}, {2, 6, 0, 0}, {4, 3, 5, 0}};
    double **pU = U;

    myfunction(&pU);
    return 0;
}

Running the code gives the following output:

10.00000
1.00000
Segmentation fault (core dumped)

Why is it that **U seemingly can only access the first row of the matrix? What should I do differently so that printf("%f\n", U[1][0]); prints the correct value?

edited Oct 09 '19 at 14:54

NathanOliver

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asked Oct 09 '19 at 14:54

A. A. Pedersen

4

Wouldn't `&pU` be a `double***` (3 stars)? – François Andrieux Oct 09 '19 at 14:55
2

`double **U` does not tell the function anything about the matrix dimensions. – Weather Vane Oct 09 '19 at 14:56
@WeatherVane What can I do instead? :/ – A. A. Pedersen Oct 09 '19 at 14:58
1

Define `int myfunction(double U[][4])` and call it with `myfunction(U);` – Weather Vane Oct 09 '19 at 14:59

Segmentation fault by using pointer to pointer

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