COMP1521 2022T2 Q4 · MIPS数组

COMP1521 2022T2 Q4 · 10 marks · MIPS
知识点: MIPS数组（MASM-A） · 次: 循环/条件 · 难度: 中等
Topic: MIPS arrays · Source: https://cgi.cse.unsw.edu.au/~cs1521/26T1/resources/practice_exams.html

一句话 / Summary: 数组中最长严格递增连续子序列长度

You have been given 22t2final_q4.s ,
a MIPS program that currently reads 10 numbers
and then prints 42 .

Your task is to modify 22t2final_q4.s
so that it is equivalent to this C program:

// Reads 10 numbers into an array
// Prints the length of the longest sequence
// of strictly increasing numbers in the array.

#include <stdio.h>

int main ( void ) {
int i ;
int numbers [ 10 ] = { 0 };

i = 0 ;
while ( i < 10 ) {
scanf ( "%d" , & numbers [ i ]);
i ++ ;
}

int max_run = 1 ;
int current_run = 1 ;
i = 1 ;
while ( i < 10 ) {
if ( numbers [ i ] > numbers [ i - 1 ]) {
current_run ++ ;
} else {
current_run = 1 ;
}

if ( current_run > max_run ) {
max_run = current_run ;
}

i ++ ;
}

printf ( "%d \n " , max_run );

return 0 ;
}

The program 22t2final_q4.c returns the length
of the longest consecutive sequence of strictly
increasing numbers.

For example:

1521 mipsy 22t2final_q4.s
1
2
3
4
5
6
7
8
9
10
10
1521 mipsy 22t2final_q4.s
1
2
3
4
5
6
7
7
8
9
7
1521 mipsy 22t2final_q4.s
1
2
3
1
2
4
5
5
6
6
4
1521 mipsy 22t2final_q4.s
10
9
8
7
6
5
4
3
2
1
1

You can use make(1) to build the provided C code:

make 22t2final_q4

You can use 1521 mipsy to run your MIPS code:

1521 mipsy 22t2final_q4.s

Your program can assume that all integers will fit in a MIPS register.

You may define and call your own functions if you wish.

Your solution must be in MIPS only.

Your solution must not be in C , C++ , Python , or any other language.

No error checking is necessary or required.

🤖 Write MIPS (mipsy). Read exactly 10 integers; print the length of the longest run of strictly increasing consecutive values (minimum 1).

Verified on the sandbox AND cross-checked against an independent oracle (official examples included).