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i am trying to calculate efficiency of a c program. I am using cclock() function to calculate the time consumed in execution of the program.
This is the code
but i am getting negative output
#include <time.h>
#include <stdio.h>
int main(){
clock_t start,end,total;
start = clock();
int i;
for(i=0;i<1000;i++)
{
printf("%d\n",i);
}
end = clock();
printf("%0.2f",(start-end)/CLOCKS_PER_SEC);
return (0);
}
Use end-start, not start-end!
start - end is subtracting the greater time from the lesser, hence why you're getting a negative value. It's like asking why you get a negative value when you subtract seven from three :-)
You need to subtract start from end.
As an aside, there's no guarantee that clock_t and CLOCKS_PER_SEC are floating point types. If they're integral types, you're likely to end up with a integer division. If you really want a floating value, use something like:
(double)(end - start) / CLOCKS_PER_SEC
There's a few other minor nigglies in your code, nothing major, but a cleaner implementation in my opinion would be along the following lines:
#include <time.h>
#include <stdio.h>
int main (void) {
clock_t duration = clock();
for (int i = 0; i < 1000; i++)
printf ("%d\n", i);
duration = clock() - duration;
printf ("%0.2f seconds\n", (double)(duration) / CLOCKS_PER_SEC);
return 0;
}
As a second aside, outputting a thousand lines is likely to take about zero seconds CPU time on modern systems. I'm assuming your payload will be a little more complex but don't be surprised if that code above takes no CPU time at all (at least to two decimal places). In fact, I had to multiply the loop by a hundred to get it up to 0.3 seconds on my system.
And don't think you can get a larger duration by using something like sleep() since that will almost certainly not use any CPU time.
printf("%0.2f",(end - start)/CLOCKS_PER_SEC);
It's simple maths. Do end - start.
The ending time is after starting time.
You should change it to printf("%0.2f",(end-start)/CLOCKS_PER_SEC);
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The following C code takes so long to compile and execute.
#include <stdio.h>
int main()
{
int max;
printf("Enter a value for max: ");
scanf("%d", &max);
for (int i=0; i<max; i++)
{
printf("%d\t", i);
}
return 0;
}
But, If I initialize the max variable with some number like
int max = 0;
The Compilation and execution is almost instantaneous. Can someone explain why?
Edit:
When I printed the value of the variable max before my input, it showed 2203648 (some garbage value). Instead of "int max = 0", if i assign
int max = 2203648;
the compilation and execution takes the same long time. But, as mentioned earlier, if i assign max say
int max = 200;
the compilation and execution is instantaneous. Does it have to do anything with the pre-assigned garbage value?
Also, this problem occurs only in windows computers, I tested with ubuntu, and the compilation and execution is instantaneous in both version of the code.
In Windows 10:
compilation and execution, as of "Enter a value for max: " appears on screen:
without variable initialization = around 8 seconds
with variable initialization = instantaneous
compiler - gcc
The scanf is failing. Check the return value.
i.e.
if (scanf("%d", &max) != 1) {
fprintf(stderr, "Unable to read max");
exit(1);
}
max is probably some large value hence the large amount of time
EDIT
The delay to see the prompt is that the printf is in a buffer and will not be displayed until the loop completes
I think variable initialization of max=0 doesn't make that great deal of a difference, but god knows why in ypur case its taking 8seconds. I think you should reinstall your GCC Compiler set your path variables correctly once more, and try the above code on a different IDE.
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I am trying to do some recursive multiplication.
When the iteration number is more than 1 million, the completion time started to go up, why?
#include <time.h>
#include <stdio.h>
float num;
unsigned long i, j;
clock_t start, end;
int main(void)
{
start = clock();
for (j = 0; j<10000000; j++){
num = 1.000001E30f;
for (i = 0; i<100; i++){
num = num * 0.999915454854432f;
if (num == 0){
printf("zero\n");
}
}
//printf("%e\n", num);
//printf("%ld\n", j);
}
end = clock();
float cpu_time_used = ((float)(end - start))/CLOCKS_PER_SEC;
printf("%f", cpu_time_used);
return 0;
}
Compiled with GCC 7.3 on Windows 10
You keep multiplying an accumulator by 0.999915454854432f, thus bringing the value closer and closer to zero. You might be getting so close to zero that it becomes a denormal representation. That may trigger slower execution in the floating point hardware and can be a source of surprising performance bloat. Just a wild guess!
See the "Performance Issues" section in the above Wikipedia page.
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I want to display the progress of my C program in percentages while it is running. The actual work in the program whose progress should be measured is confined in a loop. Here is what I tried:
int i;
int to = 100000000;
while (i++ < to) {
printf("\rPercent done: %d", (100 * i)/to);
}
Might be a dumb question, but how does one display a progress while
the program is running?
Not like this.
You have multiple issues:
Your i is uninitialized, so the program will print garbage. (Fix: -> int i = 0; instead)
Your "progress counter" will only count the progress while in the loop. As soon as Percent done: 100 will be printed, only the loop will be over.
You're printing 100 million lines to the console. Maybe think that through again.
With (i*100)/to you're hitting integer overflow about half way through, so use i / (to / 100) instead. Notice depending on the compiler the compiler could optimize that out by itself.
A little less obnoxious way would be:
#include <stdio.h>
int main (void) {
int i = 0;
int to = 100000000;
while (i++ < to) {
if (i % 1000000 == 0) {
printf("\rPercent done: %d", i / (to / 100));
}
dostuff();
}
printf("\rLoop finished");
return 0;
}
Note that this will only accurately reflect how far the program has come executing the loop. Any work before/after the loop will not be measured by this.
This only prints a console message for every full percent. Still obnoxious that you're getting 100 console messages, but nowhere near as bad as 100.000.000 (!) calls to printf. Still though, that's still a performance impact.
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i cant get the following code to run, ie it dosent give out any output
the objective is to find the sum of all primes below 2 million,
#include <stdio.h>
#include <math.h>
int is_prime(long long int i)
{
long long int n;
if(i==2)
return 1;
for(n=2;n<=sqrt(i);n++)
if(i%n==0)
return 0;
return 1;
}
int main()
{
long long int s=0,i=2;
for(i<2000000;i++;)
{
if(is_prime(i))
s=s+i;
}
printf("sum: %lli",s);
return 0;
}
You're using the for loop wrong. A for loop looks like this:
for(initialization; test expression; update)
But you wrote
for(i<2000000;i++;)
which should be
for(;i<2000000;i++)
ie, skip initialization, on each iteration test for i<2000000 and increment.
Your problem is that you are essentially factorising every integer up to two million for this. It's not that your code isn't working, it just is running very very slowly. If you attach a debugger to it you will most likely be at line
if(i%n==0)
most of the time.
Implement the Sieve of Eratostenes. Will be the most effective here.
Also your loop for(i<2000000;i++;) will run for 2^64 cycles.
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This is related to the "Project Euler Problem 14"
As I think, my logic and the code is good enough,but it may take few seconds to give the answer.
But when i run this code, the program stops when "starting_number" is around 103152(i can't remember the number exactly).
Can anybody please have a look at this code and, tell me where and what's wrong with this code.
Here's the code :
#include<stdio.h>
int starting_number;
int number_of_terms;
int j=0,k=0;
int term;
int main(){
for(starting_number=2;starting_number<1000000;starting_number++){
term = starting_number;
number_of_terms = 1;
while(1){
{
if(term%2==0){
term = term/2;
number_of_terms++;
} else if(term%2!=0){
term = 3*term + 1;
number_of_terms++;
}
}
if(term == 1) break;
}
if(j>=number_of_terms) //finding which chain is longer
j=j;
else if(j< number_of_terms) {
j= number_of_terms;
k=starting_number;
}
printf("\n%d",starting_number);
}
printf("\n%d(%d)\n",k,j);
return 0;
}
This one's mildly tricky, but your problem is here:
if(term == 1) break;
If the variable term becomes very large (as it can easily do) then it can overflow the int datatype.
When this happens term becomes negative. The C language modulus of a negative odd number is itself negative. Therefore, the end condition for your loop is never met.
Solve this problem by using a larger data type such as unsigned long long.
A less convoluted version of your code would appear as follows. Note that I have eliminated the global variables (those outside of your main function) because global variables are evil. I've replaced your infinite while-loop with a loop that uses an end condition. I've reduced duplication of code within the while-loop. I've eliminated the j=j case. Since printf is a slow function to run, I've commented out the prinft you had in the for loop, which improves the run-time significantly.
#include <stdio.h>
int main(){
int number_of_terms;
unsigned long long term;
int j=0,k=0;
for(int starting_number=2;starting_number<1000000;++starting_number){
term = starting_number;
number_of_terms = 1;
while(term!=1){
if(term%2==0)
term /= 2;
else
term = 3*term + 1;
number_of_terms++;
}
if(j<number_of_terms){ //finding which chain is longer
j = number_of_terms;
k = starting_number;
}
//printf("\n%d",starting_number);
}
printf("\n%d(%d)\n",k,j);
return 0;
}
And, indeed, using unsigned long long solves the problem.