I created two programs and I do not know which is faster

I have two generators of wps codes:

Code:

---

#include <stdio.h>

int main(int argc, char *argv[])
{
        int num=0;
        for(;num<99999999;++num)
        {
                if(num<10)
                    fprintf(stdout, "0000000%d\n", num);
                else if(num<100)
                    fprintf(stdout,"000000%d\n", num);
                else if(num<1000)
                    fprintf(stdout, "00000%d\n", num);
                else if(num<10000)
                    fprintf(stdout, "0000%d\n", num);
                else if(num<100000)
                    fprintf(stdout, "000%d\n", num);
                else if(num<1000000)
                    fprintf(stdout, "00%d\n", num);
                else if(num<10000000)
                    fprintf(stdout, "0%d\n", num);
        }
}

---

Two:

Code:

---

#include <stdio.h>

int main()
{
        char str[11];
        double num=0;
        for(int i=0; i<=99999999; i++)
        {
            snprintf(str, 11, "%.8f", num);
            for(int a=2; a<=11; a++)
                putchar(str[a]);
        putchar('\n');
            num+=0.0000001;
        }
}

---

The second code on my phone works faster.
I'm looking for volunteers who measure the speed of execution of these two codes and write here about the results. I make the program and I need the fastest solution, so I ask for your help!

a[11] is outside the bounds of the array. You should know that by now.

In the first program, you don't need the if/else if stuff. Just do this:

Code:

---

            printf("%08d\n", num);

---

The 8 is the field width, and the 0 before it says to pad the value with zeroes instead of spaces.

At any rate, it's probably somewhat pointless to worry about the difference in speed between these two programs since they are presumably both wasting the VAST majority of their time in the print statments.

If you really had to speed it up as much as possible, then maybe something like this:

Code:

---

#include <stdio.h>
#include <string.h>

#define SIZE 8

int main() {
    char s[SIZE+1] = {0};
    memset(s, '0', SIZE);

    for (;;) {
        fputs(s, stdout);
        putchar('\n');

        int i = SIZE - 1;
        while (i >= 0 && s[i] == '9')
            s[i--] = '0';

        if (i < 0)
            break;

        s[i]++;
    }

    return 0;
}

---

Quote:

---

Originally Posted by algorism

a[11] is outside the bounds of the array. You should know that by now.

---

Of course I know this, but the code works as read carefully the code, it says a<= and not a<. Running the code gives a stable generation, at least for me, so here you are wrong.

Don't confuse 'works' with 'bug free'.
It might 'work' for you today, but you can be sure that if you repeatedly assume stepping off the end of the array is benign, then you're in for pain at some point.

I made a harvester of 3 programs:

Code:

---

#include <stdio.h>
#include <time.h>
#include <string.h>

#define SIZE 8

#define MAX1 100000
#define MAX2 100000
#define MAX3 100000

int main()
{
        int start=time(NULL);
        for(int num=0; num<MAX1;++num)
        {
                printf("Generator 1: ");
                printf("%08d\n", num);
    }
    int end=time(NULL);
    int result1=end-start;
   
    start=time(NULL);
    char str[11];
        double num=0;
        for(int i=0; i<=MAX2; i++)
        {
            snprintf(str, 11, "%.8f", num);
            for(int a=2; a<=10; a++)
                putchar(str[a]);
        putchar('\n');
        printf("Generator 2: ");
            num+=0.00000001;
        }
        end=time(NULL);
        int result2=end-start;
       
        start=time(NULL);
        char s[SIZE+1] = {0};
    memset(s, '0', SIZE);
    int cnt=0;
    for (;;) {
        fputs(s, stdout);
        putchar('\n');
        printf("Generator 3: ");

        int i = SIZE - 1;
        while (i >= 0 && s[i] == '9')
            s[i--] = '0';

        if (cnt > MAX3)
            break;

        s[i]++;
        cnt++;
    }
    end=time(NULL);
        int result3=end-start;
       
        printf("\n\nGenerator 1: %d seconds\n", result1);
        printf("Generator 2: %d seconds\n", result3);
        printf("Generator 3: %d seconds\n", result3);
}

---

Performance gave me the following results:
Generator 1: 8 seconds
Generator 1: 7 seconds
Generator 1: 7 seconds

The following code is available:

Code:

---

#include <stdio.h>
#include <time.h>
#include <string.h>

#define SIZE 8

#define MAX1 99999999
#define MAX2 99999999
#define MAX3 99999999

int main()
{
        int start=time(NULL);
        for(int num=0; num<MAX1;++num)
        {
                printf("Generator 1: ");
                printf("%08d\n", num);
    }
    int end=time(NULL);
    int result1=end-start;
   
    start=time(NULL);
    char str[11];
        double num=0;
        for(int i=0; i<=MAX2; i++)
        {
            snprintf(str, 11, "%.8f", num);
            for(int a=2; a<=10; a++)
                putchar(str[a]);
        putchar('\n');
        printf("Generator 2: ");
            num+=0.00000001;
        }
        end=time(NULL);
        int result2=end-start;
       
        start=time(NULL);
        char s[SIZE+1] = {0};
    memset(s, '0', SIZE);
    int cnt=0;
    for (;;) {
        fputs(s, stdout);
        putchar('\n');
        printf("Generator 3: ");

        int i = SIZE - 1;
        while (i >= 0 && s[i] == '9')
            s[i--] = '0';

        if (cnt > MAX3)
            break;

        s[i]++;
        cnt++;
    }
    end=time(NULL);
        int result3=end-start;
       
        printf("\n\nGenerator 1: %d seconds\n", result1);
        printf("Generator 2: %d seconds\n", result3);
        printf("Generator 3: %d seconds\n", result3);
}

---

Who can execute this code and report the results?

$ gcc -Wall bar.c
bar.c: In function ‘main’:
bar.c:35:6: warning: unused variable ‘result2’ [-Wunused-variable]
int result2=end-start;
^
Mmm, you print the same result twice.

Door number 3 please Bob.

Code:

---

$ gcc -Wall -O2 bar.c
$ ./a.out | tail -3
Generator 1: 11 seconds
Generator 2: 34 seconds
Generator 3: 5 seconds
$ gcc -Wall bar.c
$ ./a.out | tail -3
Generator 1: 11 seconds
Generator 2: 33 seconds
Generator 3: 6 seconds

---

Quote:

---

Originally Posted by algorism

a[11] is outside the bounds of the array. You should know that by now.

In the first program, you don't need the if/else if stuff. Just do this:

Code:

---

            printf("%08d\n", num);

---

The 8 is the field width, and the 0 before it says to pad the value with zeroes instead of spaces.

At any rate, it's probably somewhat pointless to worry about the difference in speed between these two programs since they are presumably both wasting the VAST majority of their time in the print statments.

If you really had to speed it up as much as possible, then maybe something like this:

Code:

---

#include <stdio.h>
#include <string.h>

#define SIZE 8

int main() {
    char s[SIZE+1] = {0};
    memset(s, '0', SIZE);

    for (;;) {
        fputs(s, stdout);
        putchar('\n');

        int i = SIZE - 1;
        while (i >= 0 && s[i] == '9')
            s[i--] = '0';

        if (i < 0)
            break;

        s[i]++;
    }

    return 0;
}

---

---

Quote:

---

Originally Posted by Salem

$ gcc -Wall bar.c
bar.c: In function ‘main’:
bar.c:35:6: warning: unused variable ‘result2’ [-Wunused-variable]
int result2=end-start;
^
Mmm, you print the same result twice.

---

There's a small mistake, here's the correct code

Code:

---

#include <stdio.h>
#include <time.h>
#include <string.h>

#define SIZE 8

#define MAX1 100000
#define MAX2 100000
#define MAX3 100000

int main()
{
        int start=time(NULL);
        for(int num=0; num<MAX1;++num)
        {
                printf("Generator 1: ");
                printf("%08d\n", num);
    }
    int end=time(NULL);
    int result1=end-start;
   
    start=time(NULL);
    char str[11];
        double num=0;
        for(int i=0; i<=MAX2; i++)
        {
            snprintf(str, 11, "%.8f", num);
            for(int a=2; a<=10; a++)
                putchar(str[a]);
        putchar('\n');
        printf("Generator 2: ");
            num+=0.00000001;
        }
        end=time(NULL);
        int result2=end-start;
       
        start=time(NULL);
        char s[SIZE+1] = {0};
    memset(s, '0', SIZE);
    int cnt=0;
    for (;;) {
        fputs(s, stdout);
        putchar('\n');
        printf("Generator 3: ");

        int i = SIZE - 1;
        while (i >= 0 && s[i] == '9')
            s[i--] = '0';

        if (cnt > MAX3)
            break;

        s[i]++;
        cnt++;
    }
    end=time(NULL);
        int result3=end-start;
       
        printf("\n\nGenerator 1: %d seconds\n", result1);
        printf("Generator 2: %d seconds\n", result2);
        printf("Generator 3: %d seconds\n", result3);
}

---

Quote:

---

Originally Posted by Salem

Door number 3 please Bob.

Code:

---

$ gcc -Wall -O2 bar.c
$ ./a.out | tail -3
Generator 1: 11 seconds
Generator 2: 34 seconds
Generator 3: 5 seconds
$ gcc -Wall bar.c
$ ./a.out | tail -3
Generator 1: 11 seconds
Generator 2: 33 seconds
Generator 3: 6 seconds

---

---

I realized why the code runs so slowly. I made this code:

Code:

---

#include <stdio.h>

int main()
{
        FILE *f;
        f = freopen("out", "a", stdout);
        for(int num=0; num<1000000;++num)
                printf("%08d\n", num);
        fclose(f);
}

---

For testing, I ran like this: time ./gen
The result is 2 seconds every time you start. But maybe there is an even faster solution, it's sad if you insert a tweak generator. But on my processor MTK6572 this is a very high speed.

What?!?!?!?!?!?!?

What do you mean that you "realized why the code runs so slowly"? All you've done is decreased the number of iterations by a factor of 100!!! How is that a fair way to speed it up?

The code I wrote was designed to be faster than the code you wrote. If you think about it, it's obvious why it's faster. It completely bypasses any need to turn a binary value into decimal character output since it deals with the character representation directly.