scanf problem

**holler** · 01-22-2017

Hello,
When I select a 2 option in a menu it runs a simple calculator program where I suppose scanf() causes a problem with reading numbers I input. The second number isn't the one I input but it's 0.00 and calculation always displays a result as 0.00.. Where did I make a mistake?

Code:

void hello_world();
float calculator(op, in1, in2);

int main()
{
    float input1, input2;
    char calc_operator;
    int in;

    printf("Menu:\n");
    printf("1. Hello World!\n");
    printf("2. Calculator\n");
    printf("Enter:\n");
    scanf(" %d", &in);

    switch(in)
    {
        case 1 : hello_world();
        break;
        case 2 : printf("Calculator mode\n Enter '+' '-' '*' '/'\n");
        scanf(" %c", &calc_operator);
        printf("Enter 1 number:\n");
        scanf(" %f", &input1);
        printf("Enter 2 number:\n");
        scanf(" %f", &input2);
        calculator(calc_operator,input1,input2);
        break;
    }
    return 0;
}

float calculator(op, in1, in2)
{
    switch(op)
    {
        case '+' : printf("%.2f + %.2f = %.2f", in1, in2, in1+in2);
        break;
        case '-' : printf("%.2f - %.2f = %.2f", in1, in2, in1-in2);
        break;
        case '*' : printf("%.2f * %.2f = %.2f", in1, in2, in1*in2);
        break;
        case '/' : printf("%.2f / %.2f = %.2f", in1, in2, in1/in2);
        break;
    }
}

void hello_world()
{
    printf("Hello World!\n");
}

**Salem** · 01-22-2017

Use a compiler with more warnings enabled.

Code:

$ gcc -Wall foo.c
foo.c:3:1: warning: parameter names (without types) in function declaration
 float calculator(op, in1, in2);
 ^
foo.c: In function ‘calculator’:
foo.c:33:7: warning: type of ‘op’ defaults to ‘int’ [-Wimplicit-int]
 float calculator(op, in1, in2)
       ^
foo.c:33:7: warning: type of ‘in1’ defaults to ‘int’ [-Wimplicit-int]
foo.c:33:7: warning: type of ‘in2’ defaults to ‘int’ [-Wimplicit-int]
foo.c:37:27: warning: format ‘%f’ expects argument of type ‘double’, but argument 2 has type ‘int’ [-Wformat=]
         case '+' : printf("%.2f + %.2f = %.2f", in1, in2, in1+in2);
                           ^
foo.c:37:27: warning: format ‘%f’ expects argument of type ‘double’, but argument 3 has type ‘int’ [-Wformat=]
foo.c:37:27: warning: format ‘%f’ expects argument of type ‘double’, but argument 4 has type ‘int’ [-Wformat=]
foo.c:39:27: warning: format ‘%f’ expects argument of type ‘double’, but argument 2 has type ‘int’ [-Wformat=]
         case '-' : printf("%.2f - %.2f = %.2f", in1, in2, in1-in2);
                           ^
foo.c:39:27: warning: format ‘%f’ expects argument of type ‘double’, but argument 3 has type ‘int’ [-Wformat=]
foo.c:39:27: warning: format ‘%f’ expects argument of type ‘double’, but argument 4 has type ‘int’ [-Wformat=]
foo.c:41:27: warning: format ‘%f’ expects argument of type ‘double’, but argument 2 has type ‘int’ [-Wformat=]
         case '*' : printf("%.2f * %.2f = %.2f", in1, in2, in1*in2);
                           ^
foo.c:41:27: warning: format ‘%f’ expects argument of type ‘double’, but argument 3 has type ‘int’ [-Wformat=]
foo.c:41:27: warning: format ‘%f’ expects argument of type ‘double’, but argument 4 has type ‘int’ [-Wformat=]
foo.c:43:27: warning: format ‘%f’ expects argument of type ‘double’, but argument 2 has type ‘int’ [-Wformat=]
         case '/' : printf("%.2f / %.2f = %.2f", in1, in2, in1/in2);
                           ^
foo.c:43:27: warning: format ‘%f’ expects argument of type ‘double’, but argument 3 has type ‘int’ [-Wformat=]
foo.c:43:27: warning: format ‘%f’ expects argument of type ‘double’, but argument 4 has type ‘int’ [-Wformat=]
foo.c:46:1: warning: control reaches end of non-void function [-Wreturn-type]
 }
 ^

The two main issues being
- your function definition and declaration lack any types.
- a lot of printf/scanf calls have inconsistent type/formats.

**holler** · 01-22-2017

Thanks. Now it works.

Code:

double calculator(char op, double in1, double in2);

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