将小写的字符串转换为大写？ 在C中

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;
    }