С - Why scanf() in if() statement works incorrect?-CodePudding

I am trying to make calculator in command line application, but when I use `

int a, b, result;
printf("Enter problem:\n");

if(scanf("%i %i", &a, &b))
{
    result = a   b;
    printf("%i\n", result);
}
else if(scanf("%i*%i", &a, &b))
{
    result = a * b;
    printf("%i\n", result);
}

If I write 5 5 , I have 10, but when I 5*5 it shows 32771. If I swap those statements, when I write 5*5 , I have 25, but when I write 5 5, I have 32771 too. Thanks in advance for answer!

CodePudding user response：

if(scanf("%i %i", &a, &b))

that line makes no sense as scanf returns the number of successfully scanned items.

If you want to check if scanf succeeded

if(scanf("%i %i", &a, &b) == 2)

The rest is invalid as scanf reads the data from the input stream and the second scanf will not have the full line to scan.

char line[100];
fgets(line, 100, stdin);  // check for errors
if(sscanf(line, "%i %i", &a, &b) == 2)
{
    result = a   b;
    printf("%i\n", result);
}
else if(sscanf(line, "%i*%i", &a, &b) == 2)
{
    result = a * b;
    printf("%i\n", result);
}
else
{
    printf("Invalid Input\n");
}

CodePudding user response：

That's because scanf works differently to what you think it does.

If you write "5*5" and use the above code, the first if statement is executed. scanf scans string, parses the first number and then it hits '*' which DOES NOT match the format string. scanf exits and returns 1 as this is the number of matched strings. The first branch is executed and you are using uninitialized variable b in the addition hence the nonsense result.

If you want to parse arithmetic operations you have to write the logic yourself.

CodePudding user response：

As the other answers have pointed out, you're not using scanf correctly - if you type 5 * 5, the first scanf successfully reads 5 into a, then bails because * isn't the expected and b isn't updated. However, because you had one successful conversion and assignment, scanf returns 1 so the first branch is taken.

While this is Not The Way To Do It™, a quick-and-dirty solution would be to scan for three inputs - the two operands and the operator, then branch based on the value of the operator:

int a, b, result;
char op;

if ( scanf( "%d %c %d", &a, &op, &b ) == 3 )
{
  switch( op )
  {
    case ' ' : result = a   b; break;
    case '-' : result = a - b; break;
    case '*' : result = a * b; break;
    case '/' : result = a / b; break;
    default: fprintf( stderr, "Unrecognized operator %c - exiting\n", op ); return -1;
  }
}
else
{
  fprintf( "Badly formatted input...exiting\n" );
  return -1;
}
printf( "%d %c %d = %d", a, op, b, result );

Usually, though, parsing arithmetic expressions is a little more involved than that.