Help for code conducting a linear search

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[dm_code_snippet background=”yes” background-mobile=”no” slim=”yes” line-numbers=”yes” bg-color=”#abb8c3″ theme=”dark” language=”clike” wrapped=”yes” height=”” copy-text=”Copy Code” copy-confirmed=”Copied”]

#include<stdio.h>
#include<cs50.h>
#include<string.h>
int main(void)//to search 'a' from a string
{
string t = get_string("enter");
int m = strlen(t);
printf("number of words %i",m);
int i = 0;
    for(t[i] = 0; t[i] < m; t[i]++)
    {
        if(t[i] == 'a')
        {
        printf("found");
        }
    }
}

[/dm_code_snippet]

Help/clue appreciated why the above program fails to find ‘a’.

On trying with debugger50, it appears that the value of i not incremented. Even I tried with a sample string with first character a. But that too did not succeed in spotting ‘a’.

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Reply

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Looking at the code ? The first obvious bug I could see has to do with the condition of your for loop. In other to loop through the string you have to start from 0 to Len of string minus one . Thus (int i = 0 ; I< m ; i++)

Then you can check in the loop if t[i] = “a”

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string, here you’re including the string (arrays of character) into the loop when you’re not supposed to. So it’s:

1. For (int i =0; i<strlen(t) or m; i++) { If ( t[i] == “a”) { Printf(“found\n”) }

}

On line one you’re creating an integer i that will start counting at 0 until the strlen of t (length of the t string). Line 3 is asking each time if the “i th” character of that string is equal to the letter a. So it starts with is the zero th character an a if not go to the top of the loop and start with 1. Else if it is print found.

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