Why for loop without any working code (in third part) is there?

ajiten · 12-12-2023, 09:24 PM

In the link given here, in the main() there is twice given a for-loop, that has empty 3rd part, i.e. nothing gets executed.

Kindly help by telling why such for-loop was used twice.
The code is repeated below:

Code:

#include <stdio.h>
#include <stdint.h>
#include <stdlib.h>
#define pcx putchar_unlocked
#define gcx getchar_unlocked
 
typedef int32_t ichar;
typedef int_fast64_t fint;
 
static inline fint getfi () {
    ichar c = gcx();
    while (!isdigit (c)) c = gcx();
    fint n = 0;
    while (isdigit (c)) {
        n = n * 10 + c - '0';
        c = gcx();
    }
    return n;
}

static inline void putfi (fint n, char lc) {
    if (0 == n) {
        pcx ('0');
        if (lc) pcx (lc);
        return;
    }
    char s[24];
    fint rdi = 0;
    while (n) {
        s[rdi++] = '0' + n % 10;
        n /= 10;
    }
    while (rdi) pcx (s[--rdi]);
    if (lc) pcx (lc);
}

static inline fint min2 (fint A, fint B) {
    return (A < B ? A : B);
}
 
int main () {
    fint A [100000];
    fint R [100000];
    fint T = getfi() + 1;
    while (--T) {
        fint N = getfi();
        for (fint ai = 0; ai < N; )
        A[ai++] = N - getfi();
        for (fint ri = N-1; ri >=0; )
        R[ri--] = N - getfi();
        char nlt[100008] = {0};
        fint exit = 0;
        for (fint ai = 0; ai < N; ++ai) {
             A[ai] = A[ai] + R[ai] - N;
             if (A[ai] < 1 || A[ai] > N) {
                 exit = 1; 
                 break;
             }
            if (nlt[A[ai]]){
                exit = 1; break;
            }
            else nlt[A[ai]] = 1;
       }
       if (exit) {
            pcx('-');
            pcx('1'); 
            pcx('\n'); 
            continue; 
       }
       for (fint ai =0; ai < N; )
       putfi(A[ai++], ' ');
       pcx('\n');
      }
      return 0;
}

NevemTeve · 12-13-2023, 12:32 AM

Indentational problem, should be:

Code:

    for (fint ai = 0; ai < N; ) {
        A[ai++] = N - getfi();
    }
    for (fint ri = N-1; ri >=0; ) {
        R[ri--] = N - getfi();
    }

Incrementing ai/ri inside the loop is bad idea but it works nonetheless.

MadeInGermany · 12-13-2023, 12:39 AM

The "end action", for example ai++, can be omitted if it's done in the loop block.

Code:

        for (fint ai = 0; ai < N; )
        A[ai++] = N - getfi();

The advantage of a for loop is that you can see the "initial action", "leave condition", "end action" alltogether.
Here it is still readable because the loop body is only one line.

The normal loop would use a cpu cycle more:

Code:

        for (fint ai = 0; ai < N; ai++ )
        A[ai] = N - getfi();

pan64 · 12-13-2023, 01:12 AM

you can even use a formatter to fix issues like this: https://codebeautify.org/c-formatter-beautifier
(just copy your code into the box and check the result)

NevemTeve · 12-13-2023, 03:45 AM

> The normal loop would use a cpu cycle more
With modern compilers (meaning: since 199x) you cannot this way micro-optimize the code.

sundialsvcs · 12-13-2023, 08:26 AM

I would always use "the third part" because this what every programmer expects to see.

It's kinda like "Paris In The The Spring." Your wandering eye, "knowing" what to expect, might not have noticed the second "The" until I just mentioned it . . .

. . . and a programmer, while frantically trying to locate a bug, just might not either.

If you're lucky, they might just say "D-oh!"

But they might instead say something unprintable. So, "when you write code, think about the next person who will read it – under pressure and in a hurry."

astrogeek · 12-13-2023, 12:25 PM

Quote:

Originally Posted by ajiten

Title: Why for loop without any working code (in third part) is there?

...in the main() there is twice given a for-loop, that has empty 3rd part, i.e. nothing gets executed.

Given the thread title, it seems to me that you are saying nothing gets executed, that is the loop is not executed due to the missing third expression, so why is the loop even there.

If so, that is not correct.

All three expressions are optional, the semicolons are not, so that even with for(;;); the loop will be executed (endlessly in this case).

Quote:

Originally Posted by ajiten

Kindly help by telling why such for-loop was used twice.

Only the person who wrote it can tell you why they did it in that way.

sundialsvcs · 12-18-2023, 08:35 AM

Quote:

Originally Posted by astrogeek

Only the person who wrote it can tell you why they did it in that way.

But plenty of programmers who have to follow that person, and debug his work, will be "mightily annoyed." Yes, there is a reason why you should write your code in a way that conforms to your successor's expectations. Your successor wants above all to understand your (erroneous ...) intentions "at a glance." Because, anything else is "a waste of time," and "time is money."

Frankly, there are quite a few times over the years when I looked at a block of code, said to myself "what the f&ck is this?" Ripped the whole thing out and re-wrote it in a way that was much more obvious. (Granted: Some of that code was so old that it had been written to cope with hardware restrictions that, quite thankfully, no longer exist.)

dugan · 12-18-2023, 09:18 AM

You're asking about this loop? Which I've taken the liberty of adding brackets to?

Code:

for (fint ai = 0; ai < N;) {
    A[ai++] = N - getfi();
}

It's equivalent to this:

Code:

fint ai = 0;
while (ai < N) {
    A[ai++] = N - getfi();
}

dugan · 12-18-2023, 09:59 AM

Since this came up:

There's a website called https://godbolt.org/ that you can use to check the assembler that C/C++ code would become.

The asm produced by the following two programs is not identical. I'll leave the interpretation as an exercise to someone who knows ASM better than I do.

Code:

#include <stdio.h>

int main()
{
    for (int i = 0; i < 10;) {
        printf("%d\n", i++);
    }
    return 0;
}

Code:

#include <stdio.h>

int main()
{
    for (int i = 0; i < 10; i++) {
        printf("%d\n", i);
    }
    return 0;
}

pan64 · 12-18-2023, 12:03 PM

Quote:

Originally Posted by dugan

Since this came up:

There's a website called https://godbolt.org/ that you can use to check the assembler that C/C++ code would become.

The asm produced by the following two programs is not identical. I'll leave the interpretation as an exercise to someone who knows ASM better than I do.

Code:

#include <stdio.h>

int main()
{
    for (int i = 0; i < 10;) {
        printf("%d\n", i++);
    }
    return 0;
}

Code:

#include <stdio.h>

int main()
{
    for (int i = 0; i < 10; i++) {
        printf("%d\n", i);
    }
    return 0;
}

Yes, you are right, they are not identical.
the question is if you switch on optimization what will be the result.

ntubski · 12-18-2023, 06:48 PM

Quote:

Originally Posted by pan64

Yes, you are right, they are not identical.
the question is if you switch on optimization what will be the result.

I get the same instructions, but the order is slightly different. I guess it would be the same amount of cycles either way, but it's hard to be sure since modern CPUs are complicated.

Code:

        mov     esi, ebx
        mov     edi, OFFSET FLAT:.LC0
        xor     eax, eax
        add     ebx, 1
        call    printf

Code:

        mov     esi, ebx
        xor     eax, eax
        add     ebx, 1
        mov     edi, OFFSET FLAT:.LC0
        call    printf

There is some more code around this, but apart from this sequence it's the same for both. The .LC0 refers to "%d\n".