HackMD
Assignment1: RISC-V Assembly and Instruction Pipeline
contributed by <JordyMalone>
Quiz 1 - Problem C
Problem analysis
In this problem, we utilized three functions: fabsf, my_clz and fp16_to_fp32.
- The
fabsffunction, discussed in Problem A, calculates the absolute value of a floating-point number. Rather than using traditional arithmetic operations, it clears the sign bit through a bitwise operation to return the result. - The
my_clzfunction counts the number of leading zero bits in the binary representation of an unsigned integer, starting from the most significant bit. It then returns an integer that represents how many zero bits precede the first 1 in the binary representation of the number. - The
fp16_to_fp32function converts a 16-bit floating-point number in IEEE half-precision format to a 32-bit floating-point number in single-precision format. Themy_clzfunction is also used infp16_to_fp32.
fabsf
static inline float fabsf(float x) {
uint32_t i = *(uint32_t *)&x; // Read the bits of the float into an integer
i &= 0x7FFFFFFF; // Clear the sign bit to get the absolute value
x = *(float *)&i; // Write the modified bits back into the float
return x;
}
assembly code
fabsf:
# Assume that the input float number x is in a0
li t0, 0x7FFFFFFF # Clear the sign bit
and a0, a0, t0
jr ra
my_clz
The C code from the __bulitin_clz is mentioned in Problem C as follow:
static inline int my_clz(uint32_t x) {
int count = 0;
for (int i = 31; i >= 0; --i) {
if (x & (1U << i))
break;
count++;
}
return count;
}
assembly code
.data
argument1: .word 0x01111111 // zero: 7
argument2: .word 0x00008000 // zero: 16
argument3: .word 0x00000001 // zero: 31
newline: .string "\n"
str1: .string "The leading zero of "
str2: .string " is "
.text
main:
# Argument1 condition
lw a0, argument1
jal ra, my_clz
# Prepare to print the result
mv a1, a0
lw a0, argument1
# Call the function to print the result
jal ra, print
# Argument2 condition
lw a0, argument2
jal ra, my_clz
# Prepare to print the result
mv a1, a0
lw a0, argument2
# Call the function to print the result
jal ra, print
# Argument3 condition
lw a0, argument3
jal ra, my_clz
# Prepare to print the result
mv a1, a0
lw a0, argument3
# Call the function to print the result
jal ra, print
# Exit the program
li a7, 10
ecall
my_clz:
addi t0, zero, 0
addi t1, zero, 31 # Set i = 31
loop:
blt t1, zero, exit # i < 0 then exit
addi t2, zero, 1 # t2 assign 1U
sll t2, t2, t1 # 1U << i
and t3, a0, t2 # x & (1U << i)
bne t3, zero, exit # check if condition
addi t0, t0, 1 # count + 1
addi t1, t1, -1 # --i
jal x0, loop
exit:
mv a0, t0 # save count result into a0
jr ra
print:
mv t0, a0 # save argument to t0
mv t1, a1 # save leading zero result to t1
la a0, str1 # print string 1
li a7, 4
ecall
mv a0, t0 # print arguments
li a7, 1
ecall
la a0, str2 # print string 2
li a7, 4
ecall
mv a0, t1 # print result
li a7, 1
ecall
la a0, newline # print \n
li a7, 4
ecall
jr ra # jump back to main
Optimization
Using branchless version
#include <stdint.h>
int my_clz(uint32_t x) {
int count = 0;
if (x <= 0x0000FFFF) { // check the first 16 bits
count += 16; x <<= 16;
}
if (x <= 0x00FFFFFF) { // check the first 8 bits
count += 8; x <<= 8;
}
if (x <= 0x0FFFFFFF) { // check the first 4 bits
count += 4; x <<= 4;
}
if (x <= 0x3FFFFFFF) { // check the first 2 bits
count += 2; x <<= 2;
}
if (x <= 0x7FFFFFFF) { // check the first 1 bits
count += 1;
}
return count;
}
.data
argument1: .word 0x01111111
argument2: .word 0x00008000
argument3: .word 0x00000001
newline: .string "\n"
str1: .string "The leading zero of "
str2: .string " is "
.text
main:
# Argument1 condition
lw a0, argument1
jal ra, my_clz
# Prepare to print the result
mv a1, a0
lw a0, argument1
# Call the function to print the result
jal ra, print
# Argument2 condition
lw a0, argument2
jal ra, my_clz
# Prepare to print the result
mv a1, a0
lw a0, argument2
# Call the function to print the result
jal ra, print
# Argument3 condition
lw a0, argument3
jal ra, my_clz
# Prepare to print the result
mv a1, a0
lw a0, argument3
# Call the function to print the result
jal ra, print
# Exit the program
li a7, 10
ecall
my_clz:
addi t0, zero, 0 # Initialize count to zero
mv t1, a0 # copy argument into t1
li t2, 0x0000FFFF # t2 = 0x0000FFFF (16-bit mask)
li t3, 0x00FFFFFF # t3 = 0x00FFFFFF (24-bit mask)
li t4, 0x0FFFFFFF # t4 = 0x0FFFFFFF (28-bit mask)
li t5, 0x3FFFFFFF # t5 = 0x3FFFFFFF (30-bit mask)
li t6, 0x7FFFFFFF # t6 = 0x7FFFFFFF (31-bit mask)
# Check condition 1: if (x <= 0x0000FFFF) { count += 16; x <<= 16; }
bleu t1, t2, L1
j L2
L1:
addi t0, t0, 16 # count += 16
slli t1, t1, 16 # x <<= 16
L2:
# Check condition 2: if (x <= 0x00FFFFFF) { count += 8; x <<= 8; }
bleu t1, t3, L3
j L4
L3:
addi t0, t0, 8 # count += 8
slli t1, t1, 8 # x <<= 8
L4:
# Check condition 3: if (x <= 0x0FFFFFFF) { count += 4; x <<= 4; }
bleu t1, t4, L5
j L6
L5:
addi t0, t0, 4 # count += 4
slli t1, t1, 4 # x <<= 4
L6:
# Check condition 4: if (x <= 0x3FFFFFFF) { count += 2; x <<= 2; }
bleu t1, t5, L7
j L8
L7:
addi t0, t0, 2 # count += 2
slli t1, t1, 2 # x <<= 2
L8:
# Check condition 5: if (x <= 0x7FFFFFFF) { count += 1; }
bleu t1, t6, L9
j end
L9:
addi t0, t0, 1 # count += 1
end:
mv a0, t0
jr ra
print:
mv t0, a0 # save argument to t0
mv t1, a1 # save leading zero result to t1
la a0, str1 # print string 1
li a7, 4
ecall
mv a0, t0 # print arguments
li a7, 1
ecall
la a0, str2 # print string 2
li a7, 4
ecall
mv a0, t1 # print result
li a7, 1
ecall
la a0, newline # print \n
li a7, 4
ecall
jr ra # jump back to main
Execution state
Original version
Branchless version
Converting the loop version into a branchless version can effectively reduce both the cycle count and the number of instructions.
fp16_to_fp32
static inline uint32_t fp16_to_fp32(uint16_t h) {
const uint32_t w = (uint32_t) h << 16;
const uint32_t sign = w & UINT32_C(0x80000000);
const uint32_t nonsign = w & UINT32_C(0x7FFFFFFF);
uint32_t renorm_shift = my_clz(nonsign);
renorm_shift = renorm_shift > 5 ? renorm_shift - 5 : 0;
const int32_t inf_nan_mask = ((int32_t)(nonsign + 0x04000000) >> 8) &
INT32_C(0x7F800000);
const int32_t zero_mask = (int32_t)(nonsign - 1) >> 31;
return sign | ((((nonsign << renorm_shift >> 3) +
((0x70 - renorm_shift) << 23)) | inf_nan_mask) & ~zero_mask);
}
assembly code
.data
argument1: .word 0x3C00
argument2: .word 0xC000
argument3: .word 0x7BFF
str1: .string "\nThe FP32 value of FP16 number "
str2: .string " is "
.text
main:
lw a0, argument1
jal ra, fp16_to_fp32
mv a1, a0
lw a0, argument1
jal ra, print
lw a0, argument2
jal ra, fp16_to_fp32
mv a1, a0
lw a0, argument2
jal ra, print
lw a0, argument3
jal ra, fp16_to_fp32
mv a1, a0
lw a0, argument3
jal ra, print
li a7, 10
ecall
fp16_to_fp32:
addi sp, sp, -8
sw ra, 0(sp)
add t0, zero, a0
slli t0, t0, 16 # Extend 16 bits to 32 bits (w)
li t1, 0x80000000
and t1, t0, t1 # Extract sign bit (sign)
li t2, 0x7FFFFFFF
and t2, t0, t2 # Extract mantissa and exponent (nonsign)
add a0, t2, zero # put nonsign in a1 and pass to my_clz
jal ra, my_clz
lw ra, 0(sp)
addi sp, sp, -8
bgt t3, zero, continue # if t3 > 0 == renorm_shift > 5
addi t3, zero, 0 # else renorm_shift = 0
continue:
# inf_nan_mask
li t4, 0x04000000
add t4, t2, t4 # nonsign + 0x04000000
srli t4, t4, 8
li t5, 0x7F800000
and t4, t4, t5 # (nonsign + 0x04000000) >> 8 & 0x7F800000
# zero_mask
addi t5, t2, -1 # nonsign - 1
srli t5, t5, 31
sll t2, t2, t3 # nonsign << renorm_shift
srli t2, t2, 3 # (nonsign << renorm_shift) >> 3
li t6, 0x70
sub t6, t6, t3 # 0x70 - renorm_shift
slli t6, t6, 23 # (0x70 - renorm_shift) << 23
add t2, t2, t6 # ((nonsign << renorm_shift) >> 3) + ((0x70 - renorm_shift) << 23)
or t2, t2, t4 # t2 | inf_nan_mask
li t6, 0xFFFFFFFF
xor t5, t5, t6 # ~zero_mask
and t2, t2, t5 # t2 & ~zero_mask
or t1, t1, t2 # sign | t1
exit:
mv a0, t1
jr ra
my_clz:
addi t3, zero, 0
addi t4, zero, 31 # Set i = 31
clz_loop:
blt t4, zero, clz_exit # i < 0 then exit
addi t5, zero, 1 # t5 assign 1U
sll t5, t5, t4 # t5 = (1U << i)
and t6, a0, t5 # t6 = x & t5
bnez t6, clz_exit # check if condition
addi t3, t3, 1 # count + 1
addi t4, t4, -1 # --i
jal x0, clz_loop
clz_exit:
addi t3, t3, -5 # renorm_shift - 5
mv a0, t3 # put count result into a0
ret
print:
mv t0, a0
mv t1, a1
la a0, str1
li a7, 4
ecall
mv a0, t0
li a7, 1
ecall
la a0, str2
li a7, 4
ecall
mv a0, t1
li a7, 1
ecall
jr ra # jump back to main
Provide more tests for validations.
Undo
Use Case
LeetCode: 3011. Find if Array Can Be Sorted
Description:
You are given a 0-indexed array of positive integers nums.
In one operation, you can swap any two adjacent elements if they have the same number of set bits. You are allowed to do this operation any number of times (including zero).
Returntrueif you can sort the array, else returnfalse.
Implementation
In Problem C, we selected the my_clz function. Initially, it counts leading zeros, but we modified it to count the set bits instead.
Additionally, we defined a function called swap, which is triggered by a specific condition. The main function, canSortArray, employs the bubble sort concept to achieve its implementation.
C code
#include <stdbool.h>
int countSetbits(uint32_t number) {
int count = 0; // Initialize counter
while(number) { // Continue looping while number is not zero
count += (number & 1); // Check if the rightmost bit is '1'
// and if so , increment the counter
number >>= 1; // Shift the number right by one bit
} // to check the next bit
return count;
}
void swap(int *a, int *b) {
int temp = *a;
*a = *b;
*b = temp;
}
bool canSortArray(int* nums, int numsSize) {
bool flag;
for (int i = 0; i < numsSize; i++) {
flag = false;
for (int j = 1; j < numsSize; j++) {
if (countSetbits(nums[j - 1]) == countSetbits(nums[j])) {
if (nums[j - 1] > nums[j]) {
swap(&nums[j - 1], &nums[j]);
flag = true;
}
}
}
if (flag == 0) {
break;
}
}
for (int i = 1; i < numsSize; i++) {
if (nums[i-1] > nums[i]) {
return false;
}
}
return true;
}
Assembly code
.data
nums1: .word 8, 4, 2, 30, 15 # Declare array
nums_size1: .word 5 # Size of the array
str1: .string "True\n" # Output if sorted correctly
str2: .string "False\n" # Output if sorting fails
.text
main:
la a0, nums1 # Load the starting address of the array into a0
lw a1, nums_size1 # Load the size of the array into a1
jal ra, canSortArray # Jump to canSortArray function
la a0, nums1 # Load the starting address of the array into a0
jal ra, print # Jump to print function
li a7, 10 # End the program
ecall
canSortArray:
addi sp, sp, -4 # Allocate space on the stack
sw a0, 0(sp) # Save a0 (array address) onto the stack
addi t1, zero, 1 # Initialize i = 1
addi t0, zero, 0 # Initialize flag = 0 (indicates not sorted)
outer_loop:
lw a0, 0(sp) # Reset a0 to the starting address
addi t2, zero, 1 # Initialize j = 1
addi t0, zero, 0 # Reset flag
inner_loop:
bge t2, a1, check_flag # If j >= numsSize, jump to check_flag
lw t3, 0(a0) # Load nums[j-1]
lw t4, 4(a0) # Load nums[j]
bge t4, t3, no_swap # If nums[j] >= nums[j-1], skip swap
# Calculate the number of set bits in nums[j-1]
add t5, zero, zero # Initialize bit counter
countSetbits1:
andi t6, t3, 1 # Check the least significant bit
add t5, t5, t6 # Increment bit counter
srli t3, t3, 1 # Right shift by one
bnez t3, countSetbits1 # If t3 is not 0, continue counting
add t3, t5, zero # Store the count result in t3
# Calculate the number of set bits in nums[j]
add t5, zero, zero # Initialize bit counter
countSetbits2:
andi t6, t4, 1 # Check the least significant bit
add t5, t5, t6 # Increment bit counter
srli t4, t4, 1 # Right shift by one
bnez t4, countSetbits2 # If t4 is not 0, continue counting
add t4, t5, zero # Store the count result in t4
# If the number of set bits are not equal, continue to the next element
bne t3, t4, no_swap
# Perform the swap
lw t3, 0(a0) # Reload nums[j-1]
lw t4, 4(a0) # Reload nums[j]
sw t4, 0(a0) # Store nums[j] in nums[j-1]
sw t3, 4(a0) # Store nums[j-1] in nums[j]
addi t0, t0, 1 # Set flag = 1 to indicate a swap occurred
no_swap:
addi a0, a0, 4 # Move to the next element
addi t2, t2, 1 # j++
jal x0, inner_loop # Jump back to inner_loop for the next check
check_flag:
bnez t0, outer_loop # If flag is true, continue outer_loop
# If sorting is complete, check if all elements are in order
lw a0, 0(sp) # Reset a0 to the starting address
addi t1, zero, 1 # Initialize i = 1
isSorted:
bge t1, a1, end_true # If i >= numsSize, jump to end_true
lw t3, 0(a0) # Load nums[i-1]
lw t4, 4(a0) # Load nums[i]
blt t4, t3, end_false # If nums[i] < nums[i-1], jump to end_false
addi a0, a0, 4 # Move index by 4
addi t1, t1, 1 # i++
jal x0, isSorted # Jump back to isSorted
end_true:
addi a1, zero, 1 # Set return value to true
jr ra
end_false:
addi a1, zero, 0 # Set return value to false
jr ra
print:
beqz a1, print_false # If result is false, jump to print_false
la a0, str1 # If true, load str1
li a7, 4 # System call: print string
ecall
jr ra
print_false:
la a0, str2 # If false, load str2
li a7, 4 # System call: print string
ecall
jr ra
Execution state
