Pass pointer array to function

Question

I am trying to pass a file pointer array to a function (not sure about the terminology). Could anyone please explain the proper way to send 'in[2]'? Thank you.

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

    void openfiles (FILE **in[], FILE **out)
    {
        *in[0] = fopen("in0", "r");
        *in[1] = fopen("in1", "r");
        *out   = fopen("out", "w");
    }

    void main()
    {
        FILE *in[2], *out;

        openfiles (&in, &out);
        fprintf(out, "Testing...");

        exit(0);
    }

cnicutar · Accepted Answer · 2013-03-01 06:24:34Z

2

Try:

void openfiles (FILE *in[], FILE **out)
{
    in[0] = fopen("in0", "r");
    in[1] = fopen("in1", "r");
    *out   = fopen("out", "w");
}

And call it openfiles (in, &out);. Also, "pointer array" is ambiguous. Perhaps call it "array of FILE pointers" ?

answered Mar 1, 2013 at 6:24

cnicutar

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Comments

Grijesh Chauhan · Accepted Answer · 2013-03-01 07:50:54Z

0

You need pointer to array of FILE* type , Do as I did in below function. Also add () parenthesis like (*in) to overwrite precedence Because by default [] has higher precedence over * operator. SEE: Operator Precedence

void openfiles (FILE* (*in)[2], FILE **out){
    (*in)[0] = fopen("in0", "r");
    (*in)[1] = fopen("in1", "r");
    *out   = fopen("out", "w");
}

My example over string can be useful to understand the concept:

#include<stdio.h>
void f(char* (*s)[2]){
 printf("%s %s\n", (*s)[0],(*s)[1]);    
} 
int main(){
 char* s[2];
 s[0] = "g";
 s[1] = "ab";
 f(&s);
 return 1;
}

output:

g ab

CodePad

For OP: also read Lundin's comments to my answer Quit helpful!

edited Mar 1, 2013 at 7:50

answered Mar 1, 2013 at 7:11

Grijesh Chauhan

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6 Comments

Lundin Over a year ago

Note however that in the special case of a function parameter, FILE* (*in)[2] is equivalent to FILE* in[2], both are array pointers to an array of 2 FILE*.

Grijesh Chauhan Over a year ago

@Lundin I am bit unsure. Ok then In function parameter does char* (*s)[2] == char* s[2] ?

Grijesh Chauhan Over a year ago

@Lundin I checked here And You are CORRECT. In function parameter does char* (*s)[2] == char* s[2].. Thanks :)

Lundin Over a year ago

Function parameters are quite confusing. No matter if you type a formal array pointer declaration as you did, or if you type something that looks like an array declaration as I did, or if you merely type FILE*, it will all boil down to a pointer in the end.

Lundin Over a year ago

Though please note that the version in your answer is correct, it is the most formal and stylistically correct way of declaring such a parameter. But it can be hard to read and understand for the average programmer.

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