Creating a Calculator using command line interface and malloc,-CodePudding

I am working on malloc and command-line interface where I have to create and simple calculator. I have run the program successfully but I want to print my result from the main, not from the function and I am unable to run it from the main. Every time when I run it will show garbage value. What went wrong?

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

typedef struct {
    int num1;
    int num2;
    uint8_t *ope;
    int result;
} Calc;

void SI( Calc c) {
    if(strcmp(c.ope,"add")==0)
    {
        printf(" result is : %d\n :",c.result = (c.num1   c.num2 ));
    }
    else if(strcmp(c.ope,"sub")==0)
    {
        printf(" result is :%d\n :",c.result = (c.num1 - c.num2 ));
    }
}

int main(int argc,char *argv[]) {
    Calc *pCalc = (pCalc *) malloc(sizeof(Calc));

    pCalc ->ope = (argv[1]);
    pCalc ->num1   = atoi (argv[2]);
    pCalc ->num2   = atoi (argv[3]);
    SI(*pCalc );

    printf("result is: %d\n", pCalc ->result);  // I want to print result here

    free(pCalc );

    return 0;
}

CodePudding user response：

SI(*pCalc ); passes the value of the structure (essentially a copy) to SI, and void SI( Calc c) declares SI to have a parameter c that is initialized to the passed value. Changes to c in SI affect only the parameter c; they do not affect the structure in main.

You can change SI(*pCalc ); to SI(pCalc); so that it passes the address of pCalc instead of its value, and you can change the function declaration to void SI(Calc *c) so that its parameter c is a pointer and is initialized to the passed address. Inside SI, change c. to c->. The . operator accesses a member of a structure, whereas the -> operator accesses a member of a structure that is pointed to.

Alternatively, you could modify the return type of SI so that it returns some result—either the modify structure (return type Calc) or just the individual result (return type int). Then you would add a return statement into the function to return the value, and, in the main routine, you could use x = SI(*pCalc); to assign the returned value to x.

CodePudding user response：

You can pass pointer of pCalc to the function SI (as suggested by @EricPostpischil) and set its result there itself.

Also, added some more fixes. Please read the comments // CHANGE HERE.

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

typedef struct
{   int num1;
    int num2;
    uint8_t *ope;
    int result;
} Calc;

// CHANGE HERE - accept pointer argument
void SI(Calc* c) {
    if (c == NULL)
    {
        return;
    }
    // CHANGE HERE - use strncmp instead of strcmp
    if (strncmp(c->ope, "add", 3) == 0)
    {
        c->result = (c->num1   c->num2);
    }
    else if (strncmp(c->ope, "sub", 3) == 0)
    {
        c->result = (c->num1 - c->num2);
    }
}

int main(int argc,char *argv[]) {
    // CHANGE HERE - command line arguments validation
    if (argc != 4)
    {
        printf("Unexpected number of arguments\n");
        exit(1);
    }

    // CHANGE HERE - pCalc -> Calc
    Calc *pCalc = (Calc *) malloc(sizeof(Calc));

    pCalc ->ope = (argv[1]);
    pCalc ->num1   = atoi (argv[2]);
    pCalc ->num2   = atoi (argv[3]);
    SI(pCalc);  // CHANGE HERE - pass the pointer

    printf("result is: %d\n", pCalc->result);  // I want to print result here

    free(pCalc);

    return 0;
}