C - if/else statement

Question

I'm fairly competent with if/else statements, and this is a really old assignment that I ended up turning in partially complete. But I still wanted to know exactly why my code wouldn't work.

I want my user to input name, height, and sex. My code will then display an entire sentence that says "Name is X cm tall and is a male" or "Name is X cm tall and is a female."

When I input name and enter, it immediately then skips displays both the height AND the sex. Regardless of what I input after that, it then ends the program.

Input name: Jack Input height in cm: 180 sex(M/F): Computer $

I've been playing around with this code for a while now, but I have been stuck for a while now. Any help would be greatly appreciated. Here is my code:

#include<stdio.h>

int main() {
  char name[30];
  char sex;
  float height; 

  printf("Input name: ");
  scanf("%s", name);
  fflush(stdin);
  printf("Input height in cm: ");
  scanf("%f", &height);
  fflush(stdin);

  printf("sex(M/F): ");
  scanf("%c", &sex);
  if (sex == 'M') 
  {
    printf("%s is %f cm tall and male", name, height);
  }
  else if (sex == 'F')
  {
    printf("%s is %f cm tall and female", name, height);
  }
  printf("\n");
  return 0;
}

Worked when I compiled under c++ express.. (although I stupidly put "f" not "F" .. so didnt print the first time.) — BugFinder
– BugFinder, Commented Jul 23, 2012 at 11:58

cnicutar · Accepted Answer · 2012-07-23 11:54:12Z

5

From what I can see it only skips the sex part - which is very sad to be honest :-)).

it immediately then skips displays both the height AND the sex

Input name: Jack Input height in cm: 180 sex(M/F): Computer $

You can try this:

scanf(" %c", &sex);
       ^

The space causes scanf to eat blanks before reading a character.

answered Jul 23, 2012 at 11:54

cnicutar

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6 Comments

TheBlueCat Over a year ago

Wouldn't the value still be the same? Sorry if I'm wrong, I would have though the compiler would fill in the blanks.

cnicutar Over a year ago

@TheBlueCat scanf is very particular about its input. Some specifiers consume the space, some don't.

Some programmer dude Over a year ago

To elaborate on the reason, it's because when trying to read the sex as a single character, there is already the newline in the buffer and the scanf sets the sex to the newline. Adding a space in front of the format code tells scanf to skip whitespace (i.e. space, tabs and newlines).

TheBlueCat Over a year ago

Okay, I just tried your solution and it worked on GCC/Linux 3.4. Why does this happen, I don't understand why there needs to be a space?

Nopiforyou Over a year ago

@JoachimPileborg I intentionally used fflush(stdin); to renew the input buffer. Shouldn't that have sufficed for this issue?

|

md5 · Accepted Answer · 2012-07-23 12:01:33Z

0

fflush(stdin) is a very bad idea (undefined behavior). You should replace this call with an other function call, for example this one :

static void
clean_stdin(void)
{
    int c;
    do {
        c = getchar();
    } while (c != '\n' && c != EOF);
}

With it, it seems working.

answered Jul 23, 2012 at 12:01

md5

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2 Comments

Nopiforyou Over a year ago

Yeah, I've been told multiple times from compilers and friends to avoid using scanf and gets, but I fear I've gotten accustomed to it after being taught to use these functions.

ardent Over a year ago

@Danny, why not just use fgets?

Thomas Frei · Accepted Answer · 2012-07-23 12:02:24Z

0

You're missing an ampersand on line 9:

scanf("%s", name);

shoul be:

scanf("%s", &name);

maybe this helps

answered Jul 23, 2012 at 12:02

Thomas Frei

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3 Comments

rakeshNS Over a year ago

scanf("%s", name); is enough. No need for &

Nopiforyou Over a year ago

@TheBlueCat I am told that because the string in C is already a character array, which is a pointer value, adding & on top of that would be repetitive.

Some programmer dude Over a year ago

@TheBlueCat It's not technically wrong, since taking the address of an array is the same as using the array itself. I.e. name == &name. However, it only works for arrays, not for pointer, so if you have e.g. char *nameptr = name;, then using &nameptr will actually return the address of the pointer, not what it points to.

rakeshNS · Accepted Answer · 2012-07-23 12:02:26Z

0

You can also try this

printf("sex(M/F): ");
scanf("%s", &sex);

answered Jul 23, 2012 at 12:02

rakeshNS

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Comments

DevN · Accepted Answer · 2012-07-23 16:24:11Z

0

To read a single character from stdin you have to use getchar() instead of scanf("%c", &c) Or change the variable type of sex to char[2] and use scanf("%s", sex)

answered Jul 23, 2012 at 16:24

DevN

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