This is the code that was written in c-lisp (matrix_multiplication.sexp) :
(c-lisp
(define ((mat-cons void) (arr (ptr float)) (len int))
()
(define ((fprint float) (n float) ))
(define ((matmul_print void) (arr (ptr float)) (len int))
(declare (i int))
(for ((set i 0) (lt i len) (set i (add i 1)))
(call fprint (load (ptradd arr i)))
)
(ret)
)
;x is 10 , xout and w is 100 . d is 10 . n is 10
(define ((mat_mul void) (xout (ptr float)) (x (ptr float)) (w (ptr float)) (n int) (d int))
(declare (i int))
(for ( (set i 4) (lt i d) (set i (add i 1)) )
(declare (val float))
(set val 0.0)
(declare (j int))
(for ((set j 0) (lt j n) (set j (add j 1)))
(set val (fadd val (fmul (load (ptradd w (add (mul i n) j))) (load (ptradd x j)))))
)
(store (ptradd xout i) val)
)
(ret)
)
(define ((main void))
(declare (xout (ptr float)))
(declare (x (ptr float)))
(declare (w (ptr float)))
(declare (n int ))
(declare (d int))
(set xout (alloc float 100))
(set x (alloc float 10))
(set w (alloc float 100))
(set n 10)
(set d 10)
(call mat-cons w 100)
(call mat-cons x 10)
(call mat_mul xout x w n d)
(call matmul_print xout 100)
;(call fprint xout)
(ret)
)
)
So in the statement (for ( (set i 4) (lt i d) (set i (add i 1)) ) which is in the third line of the the function mat_mul void . We receive output for first 4 places, the next 6 positions are 'nan'. The rest of positions from 11 to 100 has shown a value.
GlowingScrewdriver
commented
4 months ago
Command executed :
This is the code that was written in c-lisp (matrix_multiplication.sexp) :
So in the statement
(for ( (set i 4) (lt i d) (set i (add i 1)) )which is in the third line of the the functionmat_mul void. We receive output for first 4 places, the next 6 positions are 'nan'. The rest of positions from 11 to 100 has shown a value.When
(set i 2), the first 2 positions has a value, the next 8 positions are 'nan', followed by values present for the next 90.This behavior is shown only within first 10 positions of the array.