can it be in programming language C, thanks 1. (20 pt) Implement a function char

Question

can it be in programming language C, thanks

1. (20 pt) Implement a function char head_tail_trim(char s1, int h, int t)i which copies the characters (except the first h character and last t characters) from the given string s1 into a dynamically created new string and returns the pointer to this new string. If t+h is greater than the length of the string, or h or t is less than 0, return NULL For example After s- head tail trim("ABCDEFG", 2, 3) s should be pointing to "CD" After s head tail trim("ABCDEFG", 0, 3 s should be pointing to "ABCD" After s -head tail trim("ABCDEFG", 3, 4) should be pointing to "" After s-head tail trim("ABCDEFG", 5, 4 s should be NULL After s= head tailtrim ("ABCDEFG", -2, 3); s should be NULL. - Suppose standard libraries are included. So, standard library functions can be used if needed. char * head tail trim (char *s1, int h, int t) (/* you can use either pointer or array notation /

Explanation / Answer

// C code

#include<stdio.h>

#include<stdlib.h>

#include <string.h>

char *head_tail_trim(char *s1, int h, int t)

{

if( (h+t) > strlen(s1) || h < 0 || t < 0 || h > t)

return NULL;

char *result = malloc(sizeof(int)*(t-h+1));

int i = 0;

while(h <= t)

{

result[i] = s1[h];

i++;

h++;

}

return result;

}

int main()

{

char *s = head_tail_trim("ABCDEFG", 2, 3);

printf("%s ",s);

s = head_tail_trim("ABCDEFG", 0, 3);

printf("%s ",s);

s = head_tail_trim("ABCDEFG", 3, 4);

printf("%s ",s);

s = head_tail_trim("ABCDEFG", 5, 4);

printf("%s ",s);

s = head_tail_trim("ABCDEFG", -2, 3);

printf("%s ",s);

return 0;

}

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