reversing only certain words of a string

Question

I have a string with the name \"Mustang Sally Bob\" After i run my code i want the string output to be like this: gnatsuM yllaS boB My approach is to count the word

Answer 1

In general the approach is incorrect.

For example an arbitrary string can start with blanks. In this case the leading blanks will not be outputted.

The last word is ignored if after it there is no blank.

The variable words does not keep the position where a word starts.

Calculating the length of a string with this loop

for(length=0;test[length] !=0&&test[length];length++);

that can be written simpler like

for ( length = 0; test[length] != '\0' ; length++ );

is redundant. You always can rely on the fact that strings are terminated by the zero-terminating character '\0'.

I can suggest the following solution

#include 

int main( void )
{
    const char *test = "Mustang Sally Bob";

    for ( size_t i = 0; test[i] != '\0'; )
    {
        while ( test[i] == ' ' ) putchar( test[i++] );

        size_t j = i;

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

        for ( size_t k = i; k != j; ) putchar( test[--k] );
    }

    return 0;
}

The program output is

gnatsuM yllaS boB

You could append the program with a check of the tab character '\t' if you like. In C there is the standard C function isblank that performs such a check.

Here is a demonstrative program that uses the function isblank. I also changed the original string literal.

#include 
#include 

int main( void )
{
    const char *test = " Mustang\tSally\tBob ";

    puts( test );

    for ( size_t i = 0; test[i] != '\0'; )
    {
        while ( isblank( ( unsigned char )test[i] ) ) putchar( test[i++] );

        size_t j = i;

        while ( test[i] != '\0' && !isblank( ( unsigned char)test[i] ) ) i++;

        for ( size_t k = i; k != j; ) putchar( test[--k] );
    }

    putchar( '\n' );

    return 0;
}

The program output is

 Mustang    Sally   Bob 
 gnatsuM    yllaS   boB