Thread: Having trouble with an if statement!

Okay, let's pretend that youjust finished typing in the last name... Your program suspends execution and waits for you to type something in...

Code:

Guybrush<ENTER>

When you press enter, your program resumes execution. It then reads in the word Guybrush into fNametemp. It does not read in the '\n' from you pressing enter, so you still have a '\n' in the buffer.

Of your first call to getchar(), which happens to be

Code:

choice1 = getchar();

The getchar() will get the '\n' that is still in the buffer. A side effect of this is that since your program doesn't have to refill the buffer, it won't prompt the user for input. The real problem, however, is that you have given choice1 the value of '\n', when you want a Y or N.

On the line after that...

Code:

getchar( );

Since you've read in the last character of the buffer, the buffer is empty, so it will wait for user input again. This next character will presumably be the 'Y' or 'N', but this isn't assined to anything, so the value is lost.

A loop can take care of this... replace

Code:

choice1 = getchar();
getchar( );

with

Code:

for (; isspace (choice1 = getchar()); );

After this line, choice1 will be the next non-whitespace character typed in.

case 'I':;
.
.
.

Question: why is there a ; after your case 'I':?

I always thought a switch was as such:

Code:

switch(something)
{
   case 'a':
   {
      blah blah;
      blah bleh;
      break;
   }
   case'b':
      break;
}

::Just had my Sociolinguistic Anthropology final this afternoon and am trying not to think too much Goddess love those minors::

> Question: why is there a ; after your case 'I':?

While it serves no purpose, the ; at the end does no harm.
You can, in C, in most places, have ; by itself and it will have no effect. a ';' just denotes the end of a statement. This is valid code:

int main ( void ) {;;;;;;;;;;;;;;;;;;;;;;;;;;; return 0;}

A ';' by itself is simply considered a 'null statement', meaining it has no effect. It _DOES_ have an (usually unwanted) effect in places like:

if( a < b );
{
printf("This will _always_ display!");
}

In places like:

for(x=0;x<10;x++);
{
printf("This only displays once!");
}

But in this case:

switch(x)
case 't': ;
{
printf("this block still executes, the ; has no adverse effect");
}

We're ok. Anyway...

Code:

case 'I':
{ 
    /**
    *** What is the purpose of these next two?
    **/
    char Y[]="Y"; 
    char N[]="N"; 

    char choice1 = 'Z'; 
    k = 5 - j; 


    printf("\nThere are currently %i spaces in the data base", k); 
    if(j<5) 
    { 
        printf("\n\nEnter the Student's first name:\t\t "); 
        scanf("%s", studalltemp[j].fNametemp); 

        printf("\ndatabase. Please conform that these details are correct (Y/N)"); 
        choice1 = getchar(); 
        getchar( ); 

        k = 0; 

        /**
        *** You may want: if( upcase(choice1) == 'Y' )
        **/
        if (choice1 == 'Y')
        { 
                printf("\n1 student record saved to file\n"); 
                strcpy(studall[j].fName, studalltemp[j].fNametemp); 
                j++; 
        } 
        else 
        /**
        *** You may want: if( upcase(choice1) == 'N' )
        **/
        if (choice1 == 'N') 
        {
                printf("Here"); 
        }
        else 
        {
                printf("%c is not a valid option.", choice1 ); 
        }
} 
break;

Try that on for size.

Quzah.

Try getche() instead of getchar()