CODEWITHSUNDEEP
cstringchar
ssd
ssd

Reputation: 2432

Convert char* string to upper

Following code is supposed to return the upper case string of the source. It works but does not convert the string. Could not figure out what was wrong.

char *StrUpper (char *s) {
    int i = 0;
    char *t = &s [i];
    while (*t) {
        if ((*t > 0x5a) && (*t < 0x7b)) t = (t - 32);
        t = &s [i++];
    }
    return (s);
}

int main () {
    printf ("%s\n", StrUpper ("lower case string"));
    return (0);
}

Upvotes: 0

Views: 169

Answers (4)

vreddy
vreddy

Reputation: 173

you might also want to change 

  t = &s[i++];

to

t =  &s[++i];

because , i++ first assigns current i value to the index then performs the increment, you end up processing the same index over again the first time.

Upvotes: 0

Vlad from Moscow
Vlad from Moscow

Reputation: 311068

First of all these two statements

if ((*t > 0x5a) && (*t < 0x7b)) t = (t - 32);
t = &s [i++];

has no sense.

This expression statement

t = (t - 32);

changes the pointer itself. And this statement

t = &s [i++];

does not change the pointer at once.

Moreover it is not clear what these magic numbers, 0x5a, 0x7b, 32 mean.

The function could be written like

char * StrUpper ( char *s ) 
{
    for ( char *p = s; *p; ++p )
    {
        if ( *p >= 'a' && *p <= 'z' ) *p -= 'a' - 'A';
    } 

    return s;
}

Here is an example of using the function

#include <iostream>

char * StrUpper ( char *s ) 
{
    for ( char *p = s; *p; ++p )
    {
        if ( *p >= 'a' && *p <= 'z' ) *p -= 'a' - 'A';
    } 

    return s;
}

int main() 
{
    char s[] = "hello";

    std::cout << s << std::endl;
    std::cout << StrUpper( s ) << std::endl;

    return 0;
}

The output is

hello
HELLO

Take into account that some coding systems do not guarantee that all lower case letters follow each other without intermediate codes. So in general this approach is not correct.

Upvotes: 0

Some programmer dude
Some programmer dude

Reputation: 409364

A string literal is a constant pointer to an array of constant characters. In other words, string literals are read only.

Trying to modify constant data leads to undefined behavior. And if your program have undefined behavior, nothing about its behavior can be trusted at all.

Upvotes: 4

MByD
MByD

Reputation: 137382

You have a few problems in your code:

  1. You forgot to dereference the string in the assignment:

    t = (t - 32);

    should be

    *t = (*t - 32);

  2. You don't check the correct range:

    if ((*t > 0x5a) && (*t < 0x7b))

    should be 

    if ((*t > 0x6a) && (*t < 0x7b))

    or, even better

    if ((*t >= 'a') && (*t <= 'z'))

  3. You go over the first character twice:

    t = &s [i++];

    should be

    t = &s [++i];

    or simply

    t++;

Upvotes: 2

Related Questions