Thread: Why doesn't it properly display the string?

The program should store a character array in reverse order then display the reversed array. I have also included in the code that will display the actual characters into the array as it loops through. So I know the characters are being stored, but why doesn't it display the entire string when I call it?

Code:

#include<stdio.h>
int main()
{
    char str[50];
    char rev[50];

    printf("Enter Desired Text: ");
    scanf ("%[^\n]%*c", str); 

    int i=0,j=0,k=0;
    while(str[i++]!='\0');
    while(i>=0)
    {
    rev[++j]=str[--i];
    /*Just Double checking
    What should go into the array*/
    printf("%c",rev[j]);
    }
    rev[j]='\0';
    printf("\n");

    printf("String: %s ",rev);
    return 0;
}

Is it just my compiler?

You are likely reversing the string twice.
You normally go halfway into a string to reverse once.

Tim S.

Originally Posted by stahta01

You are likely reversing the string twice.
You normally go halfway into a string to reverse once.

Tim S.

he does not do in place reversing - so he should copy whole string...

zerokernel,

What is the first char you are copying?

Looks like you are starting with zero character - making your resulting string empty...

Originally Posted by vart

he does not do in place reversing - so he should copy whole string...

zerokernel,

What is the first char you are copying?

Looks like you are starting with zero character - making your resulting string empty...

I just noticed something else weird with my count when it reaches the null character. Is it wrong? Lets assume my entered test was "Hello World" and I set "i" initially to be 5. How is it that my value for i reach 12?

According to what is outputed in my terminal the first character entered to the array is "d" so I'm not sure if it's wrong. Would a terminal output a null character, then continue till it gets to a letter?


I just tested my code again with a simple printf("\0"); So yes if my program was printing out null characters it wouldn't even display them on the prompt as a space or a return. So my test output for my loop ignores the null characters and then just displays the text. All I see is "dlroW olleH"

Code:

#include<stdio.h>
int main()
{
    char str[50];
    char rev[50];

    printf("Enter Desired Text: ");
    scanf ("%[^\n]%*c", str); 

    /*How is this possible?
    I should only get a value less than 12
    assuming the text entered was "Hello World"
    Why is it incrementing i wrong? Is it my scanf?
    */
    int i=5,j=0,k=0;
    while(str[i++]!='\0');
    printf("Value: %d \n",i);
    while(i>=0)
    {
    rev[++j]=str[--i];
    /*Just Double checking
    What should go into the array*/
    printf("%c",rev[j]);
    }
    rev[j]='\0';
    printf("\n");

    printf("String: %s ",rev);
    return 0;
}

you can use something like

Code:

if(rev[j] != '\0')
{
   printf("%c",rev[j]);
}
else
{
   printf("\\0");
}

Code:

    while(str[i++]!='\0');

When str[i] == '\0' you are still incrementing "i".
You need to increment "i" in the body if you don't want to count the '\0'.

Code:

    rev[++j]=str[--i];

You increment "j" before assigning a value to "rev[j]", thus you don't store anything in "rev[0]". It probably happens that "rev[0]" is 0 thus you are printing an emtpy string. Try adding

Code:

printf("%d\n", rev[0]);

after the loop to see what's actually there.

I suggest you read again how post- and pre-incrementing works.

Bye, Andreas