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Roemerdt
Roemerdt

Reputation: 147

Loop through odd numbers and 2

Is it possible in C to have a fast for/while loop that loops through the odd numbers and 2? Without using arrays.

So I'd like it to loop through {1, 2, 3, 5, 7, 9, ..}

Upvotes: 0

Views: 491

Answers (4)

0___________
0___________

Reputation: 67546

#include <stdio.h>

int main()
{
    for(unsigned x = 0; x < 10; x++)
    printf("%u%s element - %u\n",x + 1, !x ? "st" : x == 1 ? "nd" : x == 2 ? "rd" : "th", !x + x * 2 - (x >= 2));

    return 0;
}

no jumps calculating in the !x + x * 2 - (x >= 2) so no pipeline flushes.

Upvotes: 0

Richard Chambers
Richard Chambers

Reputation: 17593

Another alternative which does use an array but only a small one that is a constant size of two elements no matter how many numbers in the sequence would be:

{
    int i;
    int  iray[] = {1, 2};
    int n = 15;

    for (i = 1; i < n; i += iray[i > 2]) {
        printf (" i = %d \n", i);
        // code
    }
}

which produces:

i = 1
i = 2
i = 3
i = 5
i = 7
i = 9
i = 11
i = 13

Extending this alternative to other sequences

And this alternative can be extended to other sequences where there is a change of a similar nature. For instance if the desired sequence was

1, 2, 3, 5, 8, 11, ..

Which involves several changes in the sequence. Beginning at 1 an increment of 1 is used followed by a first increment change beginning at 3 where an increment of 2 is used followed by a second change in the sequence beginning at 5 where an increment of 3 is used, you can make the following modification.

{
    int i;
    int  iray[] = {1, 2, 3};  // increment changes
    int n = 15;

    // calculate the increment based on the current value of i
    for (i = 1; i < n; i += iray[(i > 2) + (i > 3)]) {
        printf (" i = %d \n", i);
        // code
    }
    return 0;
}

which would produce:

 i = 1
 i = 2
 i = 3
 i = 5
 i = 8
 i = 11
 i = 14

Upvotes: 0

Tim Randall
Tim Randall

Reputation: 4145

Another option

for(int i=1;;++i) // you didn't specify a limit
{
    switch(i)
    {
        default:
            if(!(i&1))continue;
        case 1:
        case 2:
            DoSomething(i):
    }
}

Upvotes: 0

klutt
klutt

Reputation: 31389

Of course. Here is a pretty straight forward way.

for(int i=1; i<N; i++) {
    if(i>3) i++;
    // Code
}

A bit more hackish variant:

for(int i=1; i<N; i+=1+(i>2)) {
     // Code
}

But I think in this case that the most readable variant would be something like:

// Code for 1 and 2

// Then code for 3,5,7 ...
for(int i=3; i<N; i+=2) {
    // Code
}

Upvotes: 3

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