Intel® oneAPI DPC++/C++ Compiler Developer Guide and Reference

Developer Guide and Reference

Contents

loop_count

Specifies the iterations for a for loop.

Syntax

#pragma loop_count
(
n
)
#pragma loop_count
=
n
or
#pragma loop_count
(
n1
[
,
n2
]
...
)
#pragma loop_count
=
n1
[
,
n2
]
...
or
#pragma loop_count
min
(
n
)
,
max
(
n
)
,
avg
(
n
)
#pragma loop_count
min
=
n
,
max
=
n
,
avg
=
n
Arguments
(n) or =n
A non-negative integer value. The compiler will attempt to iterate the next loop the number of times specified in
n
; however, the number of iterations is not guaranteed.
(n1[,n2]...) or = n1[,n2]...
Non-negative integer values. The compiler will attempt to iterate the next loop the number of time specified by
n1
 or
n2
, or some other unspecified number of times. This behavior allows the compiler some flexibility in attempting to unroll the loop. The number of iterations is not guaranteed.
min(n), max(n), avg(n) or min=n, max=n, avg=n
Non-negative integer values. Specify one or more in any order without duplication. The compiler insures the next loop iterates for the specified maximum, minimum, or average number (
n1
) of times. The specified number of iterations is guaranteed for min and max.
Description
The
loop_count
 pragma specifies the minimum, maximum, or average number of iterations for a
for
 loop. In addition, a list of commonly occurring values can be specified to help the compiler generate multiple versions and perform complete unrolling.
You can specify more than one pragma for a single loop; however, do not duplicate the pragma.
Example
The following example illustrates how to use the
loop_count
 pragma to iterate through the loop a minimum of three, a maximum of ten, and average of five times.
Example: Using the
loop_count
 pragma min(n), max(n), avg(n) 
#include <stdio.h> 
int i; 
int mysum(int start, int end, int a) {
  int iret=0; 
  #pragma loop_count min(3), max(10), avg(5)
    for (i=start;i<=end;i++)
    iret += a;
    return iret; 
} 

int main() {
  int t;
  t = mysum(1, 10, 3);
  printf("t1=%d\r\n",t);
  t = mysum(2, 6, 2);
  printf("t2=%d\r\n",t);
  t = mysum(5, 12, 1);
  printf("t3=%d\r\n",t); 
}

Product and Performance Information

1

Performance varies by use, configuration and other factors. Learn more at www.Intel.com/PerformanceIndex.