C语言代写 | CS 315 project03 – ARM assembly language
本次C语言代写汇编程序主要是完成字符串处理
project03 – ARM assembly language
Due
Requirements
Part 1a: substr The given C code contains an implementation of: char *substr_c(char *s1, char *s2); If s2 occurs in s1, the return value is a pointer to the first occurrence of s2 in s1. If s2 does not occur in s1, the return value is NULL. Part 1b: matches |
The given C code contains an implementation of: int matches_c(char *s1, char *s2); If s2 occurs in s1, the return value is the number of occurrences. If s2 does not occur in s1, the return value is 0. Your implementation of matches_s will call your implementation of substr_s. Part 2a: merge (half of merge sort) The given C code contains an implementation of: void merge_c(int a[], int i, int j, int aux[]); Merge_c copies two sorted subarrays into a final sorted array. The first subarray is a[i] to a[(i + j) / 2]. The second subarray is a[(i + j) / 2] to a[j]. Aux is temporary storage that you may use to create the sorted array, copying the final array back into a. Part 2b: merge_sort (the rest of merge sort) The given C code contains an implementation of: void merge_sort_c(int a[], int i, int j, int aux[]); Merge_sort_c takes an unsorted input array a and sorts it using a recursive merge sort algorithm. i starts at 0 and j starts at the length of the array. Rubric (80 points) Automated Tests. (10 points) Conformance to the Procedure Call Standard 1. +3useparametersandreturncodecorrectly (10 points) Code Quality
Extra credit (1 pt.)
|