Convert string of lower case to upper case? In C - c

Question

My professor gave me some exercises in C language... In one of them I have to pass a string as an argument to a function, this function will verify the existence of lower case letters in the array and convert it into upper case letter;
Actually there's a function to do such a thing, but I can't use string.h.
Does anyone have an idea to do it?
void converterup(char palavra[])
{
int i;
for(i = 0; i < 10; i++)
{
if(palavra[i] != 'a')
{
palavra[i] == 'A';
}
}
Would be something like this?

Answer 1

you need to include <ctype.h> before using function toupper, then use it like in example below (I edited your code, need to adjust it for your needs):
for(i = 0; i < 10; i++){
palavra[i] = toupper(palavra[i]);
}
this loop will convert 10 first characters to their upper ascii equivalents
or if you cannot use standard functions, you can use function like this:
char myUpperChar(char x){
const int delta = 'a' - 'A'; //'a' has ascii code 97 while 'A' has code 65
if(x <= 'z' && x >= 'a'){
x -= delta;
}
return x;
}

Answer 2

If a character is between 'a' and 'z', you could just add ('A' - 'a') to it to convert it to upper.
char input, output;
int diff = 'A' - 'a';
output = input;
if ('a' <= input && input <= 'z')
output += diff;
return output;

Answer 3

I guess your professor is expecting something more basic without external functions, like this.
char str[] = "hello";
int len = sizeof(str) / sizeof(char);
int i;
for(i = 0; i < len; i++) {
int ascii = str[i];
if(ascii >= 97 && ascii <= 122) {// 97 => 'a' and 122 => 'z' in ascii
str[i] = (char) (ascii - 32); // 32 is the ascii substraction of lower
} // and upper letters 'a' - 'A'
}
Then output would be:
HELLO

Answer 4

function will verify the existence of lower case letters in the
array and convert it into upper case letter;
I can't use string.h.
Then you have to do conversion yourself. Take a look at the ASCII chart.
Then you can notice that small and capital letters are 0x40 apart.
0x40 happens to be space ' ';
Loop through your array and convert only the small letters
arr[i] <= 'z' && arr[i] >= 'a'
remember that small and capital letters are ' ' apart.
arr[i] = arr[i] - ' ' ;
advance to next character in the array by increasing the index i++ and stop when you encounter the end of the string arr[i]=='\0'.
#include <stdio.h>
void converterup(char arr[])
{
size_t i = 0;
if(arr == NULL) return;
while(arr[i]) // loop till the '\0'; this is equivalent to `arr[i]!='\0'`
{
if(arr[i] <= 'z' && arr[i] >= 'a'){
arr[i] = arr[i] - ' ' ;
}
i++;
}
}
int main(void)
{
char str[] = "Hello World!";
converterup(str);
printf("%s",str);
return 0;
}
Test:
HELLO WORLD!

Answer 5

Check out my code guys, is it acceptable?
#include <stdio.h>
#include <stdlib.h>
#include <locale.h>
void strupr(char palavra[])
{
int i;
for(i = 0;palavra[i] > 60 && palavra[i] < 122; i++)
{
printf("%c", palavra[i] - 32);
}
}
int main(void)
{
setlocale(LC_ALL, "");
char palavra[10];
printf("Insira uma palavra maiúsculas: "); gets(palavra);
printf("Valor com conversão: ");
strupr(palavra);
return 0;
}