reading a binary file in Fortran

Question

I am trying to read a binary file consisting of signed 16 bit integers and there are exactly 51840000 of them. The code in C that accomplishes this looks like this:

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

int main()
{
    int16_t *arr = malloc(51840000*sizeof(int16_t));
    FILE *fp;
    fp = fopen("LDEM_45N_400M.IMG", "rb");
    if(fp == NULL)
    {
        printf("Error opening file\n");
        exit(1);
    }
    printf("Testing fread() function: \n\n");
    fread(arr, sizeof(*arr), 51840000, fp);
    fclose(fp);
    printf("%d \n", arr[51840000-2]);
    free(arr);
    return 0;
}

How would i read such a file in Fortran? IO in Fortran has always been very mysterious to me.

Vladimir F Героям слава · Accepted Answer · 2021-10-21 18:38:31Z

4

Use access="stream" and you can read it the same way. Declare your integers as integer(int16) and use the iso_fortran_env module.

use iso_fortran_env
integer :: ierr, n = 51840000
integer(int16) :: arr(n)

open(newunit=iu,file="LDEM_45N_400M.IMG", access="stream", status="old", action="read",iostat=ierr)
if (ierr/=0) stop "Error opening the file."
read(iu, iostat=ierr) arr
if (ierr/=0) stop "Error reading the array."
close(iu)

answered Oct 21, 2021 at 18:38

Vladimir F Героям слава

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