Sum of first and second array in a third array using C functions

Question

I have this code:

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

int inserimentoarray(int* arrayuno[], int* arraydue[]);
int calcolo(int* arrayuno[5], int* arraydue[5], int* arraytre[5]);

int arrayuno[5], arraydue[5], arraytre[5];

int main (/*int arrayuno[5], int arraydue[5], int arraytre[5]*/)
{
    int cont=0;
    inserimentoarray(arrayuno,arraydue);
    calcolo(arrayuno[5],arraydue[5],arraytre[5]);
    for (cont=0;cont<5;cont++) {
        printf("%d    +    %d     =     %d",arrayuno[cont],arraydue[cont],arraytre[cont]);
    }
    return 0;
}

void inserimentoarray(int* arrayuno[], int* arraydue[])
{
    int cont=0;
    for (cont=0;cont<5;cont++) {
        scanf("%d",&arrayuno[cont]);
        scanf("%d",&arraydue[cont]);
    }
}

void calcolo(int* arrayuno[5], int* arraydue[5], int* arraytre[5])
{
    int cont=0;
    for (cont=0; cont<5; cont++) {
        arraytre[cont]=arrayuno[cont];
    }
}

How can I make this work? It's my first using of functions in C and I don't know how to pass correctly an array from a function to another.

LastSecondsToLive · Accepted Answer · 2015-11-21 19:20:47Z

2

Here is a quick solution

void addArrays(const int s1[], const int s2[], int dest[], int n)
{
    int i;
    for(i = 0; i < n; ++i)         // iterate over the array indices
       dest[i] = s1[i] + s2[i];    // save sum of the two source arrays in destination array
}

s1 stands for source1, s2 for source2 and dest for destination. nis the number of array elements you arrays consist of.

edited Nov 21, 2015 at 19:20

answered Nov 21, 2015 at 18:31

LastSecondsToLive

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Comments

collapsar · Accepted Answer · 2015-11-21 18:52:11Z

The major insight should probably be the type equivalence of an array variable with a pointer to the base type.

Keep in mind that there is a difference when it comes to memory allocation: Declaring an array variable allocates the memory for its elements while this does not happen for a pointer decalration.

After the declaration you only pass around the variable. Imagine that th declaration gives you a handle on a memory location - in C only the programmer has any further information about what structure starts at this memory location ( this is oversimplifying but should give you a start), which is curse (maintenance!) and boon (efficiency!) at the same time. The modern perspective on software development clearly emphasizes design, reuse and maintainabiity and would thus consider it primarily a curse ...

In particular you do not specify array bounds in formal parameters of function

Keeping the structure of your program for instructional purposes,this would be a rectified version:

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


void  inserimentoarray( int* au, int* ad );
void  calcolo ( int* au, int* ad, int* at);

int arrayuno[5], arraydue[5], arraytre[5];


int main ()
{
    int cont=0;

    inserimentoarray ( arrayuno, arraydue );
    calcolo ( arrayuno, arraydue, arraytre );
    for (cont=0;cont<5;cont++) {
        printf("%d    +    %d     =     %d; ", arrayuno[cont], arraydue[cont], arraytre[cont]);
    }
    return 0;
}

void  inserimentoarray( int* au, int* ad )
{
    int cont=0;
    for (cont=0; cont<5; cont++) {
        scanf("%d",&(au[cont]));
        scanf("%d",&(ad[cont]));
    }
}

void  calcolo ( int* au, int* ad, int* at)
{
    int cont=0;
    for (cont=0; cont<5; cont++) {
        at[cont] = au[cont] + ad[cont];
    }
}

I have prepared a codepad (the scanf calls are replaced by hardcoded assignments).

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