Error in passing 2 dimensional array to function in c

Question

This is a sample of how I tried to pass the array to a function, it is not the actual function used but this one gives the same result. I get incomprehensible numbers whereas if I print the array from the main function it prints just fine. If it's something obvious, I'm sorry in advance but I'm new to using fucntions. Thanks!

I am using the call function: print(NP,EP,InputArray);

void print(int NP,int EP,double InputArray[][EP])
{
    int qw,er;
    for(qw=0;qw<NP;qw++)
    {
        for(er=0;er<EP;er++)
        {
            if(er==EP-1)
            {
                printf("%lf\n\n",InputArray[qw][er]);
            }
            else
            {
                printf("%lf\n",InputArray[qw][er]);
            }
        }
    }
}

Perhaps you meant `void print(int NP,int EP,double InputArray[NP][EP])` (notice the dimensions of `InputArray`)? — mah, Sep 10 '15 at 17:17
@mah that would make no difference ([outermost dimension is ignored](http://stackoverflow.com/a/22677793/1505939)) — M.M, Sep 10 '15 at 17:18
This part dosent seem to have any problem .Show calling function . — ameyCU, Sep 10 '15 at 17:35
This is not "a sample of how I tried to pass the array to a function". It is the function's definition. Please post a sample usage of the function to demonstrate "how I tried to pass the array to a function". — chux - Reinstate Monica, Sep 10 '15 at 17:35
@melpomene Why though? If it is indeed wrong then it must be it cause the output resembles that of using the wrong % to refer to it. — Michael K, Sep 10 '15 at 19:02
@MichaelK In C89/C90 `%lf` was simply undefined. In C99 it was defined to be equivalent to `%f`, so there's still no point in using it. — melpomene, Sep 10 '15 at 19:10

score -1 · Answer 1 · answered Sep 10 '15 at 22:25

-1

Turns out this syntax doesn't work for codeblocks. Tried DevC++ and works just fine.

answered Sep 10 '15 at 22:25

Michael K

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Error in passing 2 dimensional array to function in c

1 Answers1