"expected expression before 'int'" error

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"expected expression before 'int'" error

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#1 by (1 votes)
#2 by (1 votes)

0

I was following this example, but it was done in Windows, so I used <conio.h> . When I try to run on Linux I get 2 errors: one in *h=int(tempo); , I do not understand why, and the other in getch . How can I replace them?

#include <stdio.h>

void horario(float tempo, int *h, int *m)
{
    *h=int(tempo);
    *m=(tempo-*h)*60;
}

int main ()
{
    int hora,minuto;
    char resp;
    float t;

    do
    {
        printf("Digite o horario na forma centesimal: ");
        scanf("%f",&t);
        horario(t,&hora,&minuto);
        printf("Horario: %02d:02%d\n",hora,minuto);
        printf("Quer calcular outro horario? (S/N):");
        resp=getch();
    } while (resp=='S');
    getch();
    return 0;
}

c

asked by anonymous 25.03.2014 / 00:12

2 answers

1

To cast, use the desired type in parentheses followed by

(int)tempo // cast de tempo para tipo int: correcto
int(tempo) // incorrecto: aparenta ser a chamada da funcao int com argumento tempo

You can replace getch() with getchar() (prototype in <stdio.h> that you included).

25.03.2014 / 00:16

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score 1 · Accepted Answer

You had two problems:

Typecast badly done.

getchar instead of getch

void horario(float tempo, int *h, int *m){
    *h=(int)tempo;
    *m=(tempo-*h)*60;
}

int main(){
    int hora,minuto;
    char resp;
    float t;
    do
    {
        printf("Digite o horario na forma centesimal: ");
        scanf("%f",&t);
        horario(t,&hora,&minuto);
        printf("Horario: %02d:02%d\n",hora,minuto);
        printf("Quer calcular outro horario? (S/N):");
        getchar();
        resp = getchar();
    } while (resp=='S');
    getchar();
    return 0;
}