syntax problem 3

[chopstick]

this is part of the code:

while((ch=getchar())!='#')
{
if(ch=='.')
{
putchar('!');
count++;
}
if(ch=='!')
{
printf("!!");
count++;
}
else
putchar(ch);
}
I thought the "else" is associated with the second "if", but when it encounters "." it replaces it with "!", and it also prints ".", that means it goes to the last "putchar(ch); ". why is this?

 

[cpjust]

If ch == '.' then the first if block gets executed. Since the next if statement isn't an else if, it always gets evaluated. Since (ch == '!') is false, the else block gets executed instead.

 

[xwb]

Alternatively, put a continue; after the count++ in the first if statement.

 

[Salem]

1. Please use [code][/code] tags when posting code.

2. if(ch=='!')
Change this to
else if(ch=='!')

It should do what you want then

while((ch=getchar())!='#')
{                          
  if(ch=='.')
  {
    putchar('!');
    count++;
  }
  else if(ch=='!')
  {
    printf("!!");
    count++;
  }
  else
  {
    putchar(ch);                     
  }
}

--