如何在C中将字符串（char *）转换为大写或小写

Question

I have a struct:

typedef struct entry {
char *surname;
int house_no;
char *postcode;
} BEntry;

and a function to convert strings to upper case:

void toUpper(char *str){
    while (*str != '\0')
    {
        *str = toupper(*str);
        str++;
    }
}

and in my main function I assign values to the struct members and want to convert the surname to upper case:

mentry->surname = "bob";
mentry->house_no = 17;
mentry->postcode = "GK116BY";
toUpper(me->surname);

What is the correct way to convert a string to upper case by passing a char pointer to a function like this? My program is returning a segmentation fault. Any help is most appreciated, thanks.

Answer 1

Your toUpper() implementation should work fine. However, you should pay attention to the warnings from your compiler (or turn up the warning levels if you didn't get any -- you should see something when you assigne a string literal to a char * like that). String literals are const , ie they cannot be modified. When you try to write to them, this will cause the segmentation fault you are seeing.

You need something like:

mentry->surname = malloc(4);
strcpy(mentry->surname, "bob");

or the more convenient, but not part of the C standard way:

mentry->surname = strdup("bob");

And of course, be sure to call free() later, either way.

Answer 2

String literals like "Hello, World!" are not writeable. Either duplicate the string literal:

hw = strdup("Hello, World!");

or declare a char[] variable and initialize it to a string literal. As a special case, string literals used this way are writeable:

char hw[] = "Hello, World!";

Answer 3

The easiest way of converting char * to lowercase or uppercase in c, is by using strdup

#include <string.h>

char    *to_uppercase(char *s)
{
    int i = 0;
    char    *str = strdup(s);

    while (str[i])
    {
        if (str[i] >= 97 && str[i] <= 122)
            str[i] -= 32;
        i++;
    }
    return (str);
}