getchar()

I have some code

Code:

        int x;
        printf("1.  Quit\n");
        printf("2.  Add number\n");
        x=getchar();
        if(x=='2'){

At first the 2 wasn't in character constant notation, but the if statement never worked. I printed %d of x and it came out 50. Why do I have to do this? getchar always returns an int, and I have never had this problem before?

Originally Posted by linuxdude

I have some code

Code:

        int x;
        printf("1.  Quit\n");
        printf("2.  Add number\n");
        x=getchar();
        if(x=='2'){

At first the 2 wasn't in character constant notation, but the if statement never worked. I printed %d of x and it came out 50. Why do I have to do this? getchar always returns an int, and I have never had this problem before?

What do you expect to get from the following:

Code:

#include <stdio.h>
int main()
{
  int x;
  
  printf("1.  Quit\n");
  printf("2.  Add number\n");
  x=getchar();
  printf("x = %c (%d decimal), (0x%02x hex)\n", x, x, x);
  if(x=='2'){ 
    printf("It passed the =='2' test.\n");
  }
  else{
    printf("It didn't pass the =='2' test.\n");
  }

 if (x == 50) {
    printf("It passed the ==50 test.\n");
  }
  else {
    printf("It didn't pass the ==50 test.\n");
  }

  return 0;
}

What do you get?

What's the problem?

Regards,

Dave

I see what it passed, but I dont' know why it doesn't pass the 2. getchar() returns an int I have an int, there shouldn't be a need for a conversion right?

Originally Posted by linuxdude

I see what it passed, but I dont' know why it doesn't pass the 2. getchar() returns an int I have an int, there shouldn't be a need for a conversion right?

The int returned from getchar() when you press the '2' key has a value of 50 decimal. By assigning the integer value 50 to the char variable x, x now has the value 50 decimal.

When you tell printf to print x as a %c, it converts to the ascii printable character '2'. When you tell printf to print it as a %d, it prints the decimal value. When you tell printf() to print it as a %x, it prints the hex value.

When you compare x with '2', it passes, since '2' has the decimal value 50 and so does x.

When you compare x with 50, it passes, since x has the decimal value of 50

("Bits is bits.")

Dave

hmm I though all characters from the keyboard were ints.

Originally Posted by linuxdude

hmm I though all characters from the keyboard were ints.

getchar() returns an int.

If you press the key '2', getchar() returns an int with value 50 decimal.

In the C language, the notation '2' indicates a "character constant" whose value is the ascii representation of '2', which happens to be 0x32, or decimal 50.

If you set a char equal to an int, the conversion is made automatically in C by chopping off the upper bits. Since the upper bits of 50 decimal are all zero, the conversion doesn't seem to change anything.

Forgetting getchar() for the moment, try this:

Code:

#include <stdio.h>
int main()
{
  int xx;
  char cc;

  xx = 0x1234;
  cc = xx;
  printf("xx = %d (decimal), cc = %d (decimal)\n", xx, cc);
  printf("xx = %x(hex), cc = %x(hex)\n", xx, cc);

  return 0;
}

Dave

Originally Posted by linuxdude

I have some code

Code:

        int x;
        printf("1.  Quit\n");
        printf("2.  Add number\n");
        x=getchar();
        if(x=='2'){

At first the 2 wasn't in character constant notation, but the if statement never worked. I printed %d of x and it came out 50. Why do I have to do this? getchar always returns an int, and I have never had this problem before?

Are you confusing it with this?

Code:

scanf("%d", &x);

Originally Posted by linuxdude

hmm I though all characters from the keyboard were ints.

Like 'a', 'b', 'c', '1', '2', '3' which all have integer values?

Originally Posted by linuxdude

hmm I though all characters from the keyboard were ints.

Characters from the keyboard are whatever the function that gets them gives you.

When you press the '2' key, getchar() returns an int that has the value of the ascii representation of '2'. Some other function could give you something else.

Dave