Thread: Some questions regarding old C code

Hey all
I have some trouble understanding the old goto.c ( see also goto man page V6 Thompson Shell Port - Manuals - GOTO(1) ) from this source code:

Code:

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int offset 0; 
  
main(argc, argv) 
char *argv[]; 
{ 
        extern fin; 
        char line[64]; 
  
        if (argc<2 || ttyn(0)!='x') { 
                write(1, "goto error\n", 11); 
                seek(0, 0, 2); 
                return; 
        } 
        seek(0, 0, 0); 
        fin = dup(0); 
  
loop: 
        if (getlin(line)) { 
                write(1, "label not found\n", 16); 
                return; 
                } 
        if (compar(line, argv[1])) goto loop; 
        seek(0, offset, 0); 
} 
  
getlin(s) 
char s[]; 
{ 
        int ch, i; 
  
        i = 0; 
l: 
        if ((ch=getc())=='\0') return(1); 
        if (ch!=':') { 
                while(ch!='\n' && ch!='\0') 
                        ch = getc(); 
                goto l; 
                } 
        while ((ch=getc())==' '); 
        while (ch!=' ' && ch!='\n' && ch!='\0') { 
                s[i++] = ch; 
                ch = getc(); 
                } 
        while(ch != '\n') 
                ch = getc(); 
        s[i] = '\0'; 
        return(0); 
} 
  
compar(s1, s2) 
char s1[], s2[]; 
{ 
        int c, i; 
  
        i = 0; 
l: 
        if(s1[i] != s2[i]) return(1); 
        if (s1[i++] == '\0') return(0); 
        goto l; 
} 
  
getc() 
{ 
  
        offset++; 
        return(getchar()); 
}

Line 1-17: Everything ok.
My problem starts with getlin(line).
I try to comment as far as I understood:

Code:

getlin(s) 
char s[]; 
{ 
        int ch, i; 
  
        i = 0; 
l: 
// if we reach the end of the script, we return with 1 
        if ((ch=getc())=='\0') return(1); 
//If the first char is not a ':' we iterate char by char thorugh the whole line
//we goto l and check the next line for a starting ':'
        if (ch!=':') { 
                while(ch!='\n' && ch!='\0') 
                        ch = getc(); 
                goto l; 
                } 
// there is a ':' as the first char in a line found
// we skip all the spaces in front of the label   
        while ((ch=getc())==' '); 
// Now we copy just the label to s  
        while (ch!=' ' && ch!='\n' && ch!='\0') { 
                s[i++] = ch; 
                ch = getc(); 
                } 
// Question 1
        while(ch != '\n') 
                ch = getc(); 
        s[i] = '\0'; 
// Question 2
        return(0); 
}

@Question 1
Why would you do this while loop?
What would happen if we don't run this loop?

@Question 2
Why do we return with 0? We found a line, which starts with ':' and a label.
If we return 0, we would go into

Code:

 if (getlin(line)) { 
                write(1, "label not found\n", 16); 
                return; 
                }

altough we did find a label.

I hope you can help me

Thank you grumpy

1)
Imagine this script:

Code:

...
goto end
...
: end

Now let's look at this part of the code:

Code:

while (ch!=' ' && ch!='\n' && ch!='\0') {
s[i++] = ch;
ch = getc();
}
while(ch != '\n')                 
ch = getc();         
s[i] = '\0'; 

We would exit the first while-loop with ch='\0', because ": end" ist the last line of our script. And now we would enter the second while-loop.
But how will we exit the second while-loop? ch will never be '\n' because we already reached the end of our script?

2)

Return value of zero indicates a label was found. Return value of 1 indicates no label found.

Okay, I understand that. But then it should be:

Code:

if (!getlin(line)) 
{
write(1, "label not found\n", 16);
return;                 
} 

Shouldn't it?