Return string in C for manipulation outside the function in which it was declared

Question

I must develop a calculator that reads strings in the algebraic form of the complex number operation. I need to manipulate the "main" vector outside the function in which it was declared (receives). How to proceed? Should pointers be used? How to use them?

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

int tam_max = 256;

int recebe(){

    int i = 0;
    int erro = 0;
    char principal[tam_max];
    char real1[tam_max];

    setbuf(stdin, NULL);
    fgets(principal, tam_max, stdin);

    for (i = 0; i < strlen(principal); i++){

        if (principal[0] == '\n'){
            erro++;
        } else if ( principal[i] == '+' || principal[i] == '-' ||
            principal[i] == '*' || principal[i] == '/' ||
            principal[i] == '=' || principal[i] == '^' ||
            principal[i] == 'i' || principal[i] == 'p' ||
            principal[i] == '0' || principal[i] == '1' ||
            principal[i] == '2' || principal[i] == '3' || 
            principal[i] == '4' || principal[i] == '5' || 
            principal[i] == '6' || principal[i] == '7' ||
            principal[i] == '8' || principal[i] == '9' || principal[i] == '\n'){
            erro == 0;
        } else {
            erro++;
        }
    } 
    return erro;
}

int validaDados(){

    int verifica = 0;
    verifica = recebe();

    if (verifica > 0){
        printf("\nCaracteres invalidos inseridos. Por favor, tente novamente.\n\n");
        validaDados();
    }

    if (verifica == 0){
        divide();
    }
}

int divide(){
    printf("OK ate aqui\n");
}

int main(){

    validaDados();

}

Answer 1

The correct is to always allocate the necessary memory where you need it and pass this address to anyone who will manipulate it. In exercises the allocation of a array in the stack it’s not usually a problem. In more complex applications you may want to make dynamic allocation in heap.

The question does not clearly state how it will use so I will try something here creating and using the string in validaDados() and pass it as argument (as is passed like a pointer everything that is changed in it will be reflected in the original variable passed as argument.:

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

#define tam_max 256

int recebe(char principal[tam_max]) {
    int erro = 0;
    setbuf(stdin, NULL);
    fgets(principal, tam_max, stdin);
    for (int i = 0; i < strlen(principal); i++) {
        if (principal[0] == '\n') {
            erro++;
        } else if ( principal[i] == '+' || principal[i] == '-' ||
            principal[i] == '*' || principal[i] == '/' ||
            principal[i] == '=' || principal[i] == '^' ||
            principal[i] == 'i' || principal[i] == 'p' ||
            principal[i] == '0' || principal[i] == '1' ||
            principal[i] == '2' || principal[i] == '3' || 
            principal[i] == '4' || principal[i] == '5' || 
            principal[i] == '6' || principal[i] == '7' ||
            principal[i] == '8' || principal[i] == '9' || principal[i] == '\n') {
            erro = 0;
        } else {
            erro++;
        }
    } 
    return erro;
}

void divide() {
    printf("OK ate aqui\n");
}

void validaDados() {
    int verifica = 0;
    char principal[tam_max];
    verifica = recebe(principal);
    printf("%s\n", principal);
    if (verifica > 0) {
        printf("\nCaracteres invalidos inseridos. Por favor, tente novamente.\n\n");
        validaDados();
    }
    if (verifica == 0) {
        divide();
    }
}

int main() {
    validaDados();
}

Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.

I got myself into some other trouble. But I left other things that could get better. I do not know if the code does what you want, but there is the solution requested in the question.

Answer 2

Here are some options for your case:

Option 1 - The client allocates a buffer and passes its pointer as parameter to be filled in internally by the function:

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

#define MAXBUF  (100)

char * obter_msg( char * msg, size_t tam )
{
    strncpy( msg, "Ola Mundo!", tam );
    return msg;
}

int main( void )
{
    char msg[ MAXBUF + 1 ] = {0};
    obter_msg( msg, MAXBUF );
    printf("%s\n", msg );
    return 0;
}

/* OU... */

int main( void )
{
    char * msg = (char*) malloc( (MAXBUF + 1) * sizeof(char) );
    obter_msg( msg, MAXBUF );
    printf("%s\n", msg );
    free(msg);
    return 0;
}

Option 2 - The function has an internal static buffer and always returns its pointer to the client:

#include <stdio.h>
#include <string.h>

#define MAXBUF  (100)


char * obter_msg( void )
{
    static char msg[ MAXBUF + 1 ] = {0};
    strncpy( msg, "Olah Mundo!", MAXBUF );
    return msg;
}

int main( void )
{
    printf("%s\n", obter_msg() );
    return 0;
}

Option 3 - The function dynamically allocates a buffer and returns its pointer to the client (which in turn releases the buffer after use):

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

#define MAXBUF  (100)

char * obter_msg( void )
{
    char * msg = (char*) malloc( (MAXBUF + 1) * sizeof(char) );
    strncpy( msg, "Olah Mundo!", MAXBUF );
    return msg;
}

int main( void)
{
    char * msg = obter_msg();
    printf("%s\n", msg );
    free(msg);
    return 0;
}

I hope I’ve helped!