Why does loop work this way? - c

Question

I created a program. It works as I wanted it to work, but I don't understand why.
This is the function.
void LiteraMajuscula(char *str)
{
int i;
i = 0;
while (str[i] != '\0')
{
if (i == 0 && str[i] >= 'a' && str[i] <= 'z')
str[i] -= 32;
if (str[i-1] == ' ' && str[i] >= 'a' && str[i] <= 'z')
str[i] -= 32;
i++;
}
}
It is supposed to make the lowercase character into an uppercase one each time there is a space and only the first one and if the first character from a string is lowercase, make it an uppercase.
The only thing i dont understand is the str[i-1]. I tried str[i], but it doesn't change anything, and str[i-2] changes the second letter into and uppercase instead of the first one. Why is that?

Answer 1

str[i] -= 32; will convert the character at index i. The reason the second if statement uses str[i-1] == ' ' in its check is so it knows whether the current character (at i) is just after a space (at i - 1). The reason it converted the second character of words when you changed it to str[i-2] == ' ' is that you changed it so it converted a character if it (at i) was two characters after a space (at i-2).
As noted in the comments, the code there has undefined behavior, because there's nothing preventing the str[i-1] == ' ' check when i is 0, so str[i-1] would be accessing the character before where str is pointing.
Separately, since most of the conditions and logic are duplicated between the two ifs, this is where you'd use || (logical OR):
while (str[i] != '\0')
{
if ((i == 0 || str[i-1] == ' ') && str[i] >= 'a' && str[i] <= 'z')
str[i] -= 32;
i++;
}
(Note the () around the ||.)
|| short-circuits, so when i is 0 and the first operand is true, the second operand (str[i-1] == ' ') is never evaluated and so you avoid undefined behavior.

Answer 2

The body of the loop is incorrect
while (str[i] != '\0')
{
if (i == 0 && str[i] >= 'a' && str[i] <= 'z')
str[i] -= 32;
if (str[i-1] == ' ' && str[i] >= 'a' && str[i] <= 'z')
str[i] -= 32;
i++;
}
When i is equal tp 0 then in this condition
if (str[i-1] == ' ' && str[i] >= 'a' && str[i] <= 'z')
^^^^^^^^
there is an attempt to access memory beyond the string. The loop should be rewritten at least like
while (str[i] != '\0')
{
if ( ( i == 0 || str[i-1] == ' ' ) && str[i] >= 'a' && str[i] <= 'z')
str[i] -= 32;
i++;
}
That is the condition checks whether the first character of the string is an alpha character or if it is not the first character of the string whether the previous character is space and the current character is an alpha character.
Take into account that the function will not work with EBCDIC characters. It is better to use standard C function toupper declared in header <string.h>. For example
while (str[i] != '\0')
{
if ( i == 0 || str[i-1] == ' ' )
{
if (isalpha((unsigned char)s[i]) ) s[i] = toupper((unsigned char)s[i]);
}
i++;
}
I think it is desirable also to check whether the previous character is the tab character. Also such string functions usually return the target string.
I would write the function the following way
#include <stdio.h>
#include <ctype.h>
char * LiteraMajuscula( char *s )
{
for (size_t i = 0;
s[i += strspn(s + i, " \t")] != '\0';
i += strcspn( s + i, " \t" ))
{
if (isalpha((unsigned char)s[i]) ) s[i] = toupper((unsigned char)s[i]);
}
return s;
}
int main(void)
{
char s[] = "hello,\tworld!";
puts(s);
puts(LiteraMajuscula(s));
return 0;
}
Its output is
hello, world!
Hello, World!

Answer 3

The other answers posted have explained what was wrong with the code.
Here is an alternative way that the function can be rewritten to be more efficient and more portable, with the use of functions from ctype.h:
void do_some_incorrect_capitalization (char str[])
{
if(*str == '\0')
{
return ;
}
*str = toupper(*str); // special case, first letter in the string
str++;
while(*str != '\0')
{
if(isspace(*(str-1)))
{
*str = toupper(*str);
}
str++;
}
}