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function for four basic arithmetic ...

attisattis Posts: 25Member
Dear all,

I have a half-ready program to use the four basic (add, subtract, multiply, divide) arithmetic . I would need a little support to make it work.

Any help is appreciated,

attis

[code]
#include

int calcInt(int nOne, int nTwo, int nflag)
{
int nRet = 1;

if(1 == nflag){
printf(" %d
", nOne+nTwo);
nRet = 1;
}else if(2 == nflag){
printf(" %d
", nOne-nTwo);
nRet = 1;
}else if(3 == nflag){
printf(" %d
", nOne*nTwo);
nRet = 1;
}else if(4 == nflag){
printf(" %d
", nOne/nTwo);
nRet = 1;
}else{
nRet = 0;
}
return nRet;
}

int main(void)
{
int nRet = 0;

int calcInt = 1;
int nOne = 10;
int nTwo = 5;

if( 0 == calcInt){
printf("Failed!
");
}

return nRet;
}


[/code]

Comments

  • BitByBit_ThorBitByBit_Thor Posts: 2,444Member
    : Dear all,
    :
    : I have a half-ready program to use the four basic (add, subtract,
    : multiply, divide) arithmetic . I would need a little support to make
    : it work.

    I made a change in blue

    [code]

    int main(void)
    {
    int nRet = 0;

    [color=Green]/* You need to learn about how you pass parameters
    to a function :) You don't need those: */
    /* int calcInt = 1;
    int nOne = 10;
    int nTwo = 5; */[/color]

    if( 0 == [color=blue]calcInt(10, 5, 1)[/color] ){
    printf("Failed!
    ");
    }
    return nRet;
    }
    [/code]

    And if I may make a suggestion (and if I don't I'm going to make it anyway :P): change "int nFlag" parameter to "char operator".
    And then use '+', '-', '*' and '/' instead of 1-4:
    [code]
    int calcInt(int nOne, int nTwo, char operator)
    {
    switch (operator)
    {
    case '+':
    [color=Green]/* add number */[/color]
    break;
    case '-':
    ...
    default:
    /* bad operator */
    }
    }
    [/code]
    Best Regards,
    Richard

    The way I see it... Well, it's all pretty blurry
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