Function reversing a string changes string to empty, rather than reverse

Question

I am not allowed to use any headers, and I want to create code, that takes a string input and modifies it to its reverse. There are no compile-time errors, however during run-time it does change the string to empty, rather than reverse.

int str_size(char str[])
{
    int i=0;
    while (str[i]!='\0')i++;
    return i;
}

void str_rev(char str[])
{
    char buff;
    int size=str_size(str);

    if(size==0 || size==1){}
    else
    {
        for (int i = 0; i < (size/2)-1; i++)
            {
                buff=str[i]; 
                str[i]=str[size-i];
                str[size-i]=buff;
            }
    }
}

str_rev is called in main(void).

The terminal message:

iplayzed@iplayzed-Lenovo-ideapad-330-15IKB:~/Egyetem/Progalap gyak 1.fv/3_hazi$ gcc -Wall -O2 -static -o my my.c
my.c: In function ‘main’:
my.c:30:5: warning: ignoring return value of ‘scanf’, declared with attribute warn_unused_result [-Wunused-result]
     scanf("%s",str);
     ^~~~~~~~~~~~~~~
iplayzed@iplayzed-Lenovo-ideapad-330-15IKB:~/Egyetem/Progalap gyak 1.fv/3_hazi$ ./my
asdf
String length: 4

The original string:"asdf"
The reversed string:""

What is the cause of it changing to blank?

dash-o · Accepted Answer · 2019-10-05 19:11:40Z

2

The indices for the reversal are a little off by 1.

Consider size=4, then for the first iteration (i=0), code will execute:

buff = str[0]
str[0] = str[4-0] ;
str[4] = buff ;

Recall that str[4] is the terminating NUL, so that the code will insert NUL into the first position, therefore an empty string is returned.

This can be easily fixed by:

buff=str[i]; 
str[i]=str[size-i-1];
str[size-i-1]=buff;

answered Oct 5, 2019 at 19:11

dash-o

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1 Comment

Weather Vane Over a year ago

Yes, there was "empty string" because the NUL terminator was being copied to the front of the string.

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