Is it applicable to pass pointer to character (after assigning string to it) to method and modify this string after without using array and why not?

Question

I want to pass string to a c function using pointer to char and modify it and it gave me segmentation fault. I don't know why ? Note*: I know I can pass the string to array of character will solve the problem

I tried to pass it to array of character and pass to function the name of array and it works , but I need to know what the problem of passing the pointer to character.

void convertToLowerCase(char* str){

    int i=0;
    while(str[i] != '\0')
    {
        if(str[i]>='A'&& str[i]<='Z'){
            str[i]+=32;
        }
        i++;
    }
}

int main(void){

    char *str = "AHMEDROSHDY";
    convertToLowerCase(str);
}

I expect the output str to be "ahmedroshdy", but the actual output segmentation fault

@gsamaras how can you clarify more ,this is not const char* ? — ahmed roshdy, May 12 '19 at 13:21
@xing I know that but I need to know why when I use char str* , it give this error .. what happens in memory and why this is consider wrong need to understand ? — ahmed roshdy, May 12 '19 at 13:23
The string literal will still be in read-only memory (not necessarily but practically always) so you can’t change it. — Sami Kuhmonen, May 12 '19 at 13:23
@SamiKuhmonen why str become in read-only , it became as const char* str and I didn't do that to become read-only can you clarify more pls ? — ahmed roshdy, May 12 '19 at 13:26
@xing ok but why this happened it is not const char * str. and I can do str[0]= any character. right ? — ahmed roshdy, May 12 '19 at 13:29
@xing that is right . the str[0] = 'a'; is not possible because it seems the char* is read-only so any char* is just read-only ? — ahmed roshdy, May 12 '19 at 13:47

gsamaras · Accepted Answer · 2019-05-12T14:21:59.590

3

This (you had char str* which is a syntax error, fixed that):

char *str = "AHMEDROSHDY";

is a pointer to a string literal, thus it cannot be modified, since it is stored in read-only memory.

You modify it here str[i]+=32;, which is not allowed.

Use an array instead, as @xing suggested, i.e. char str[] = "AHMEDROSHDY";.

edited May 12 '19 at 14:21

answered May 12 '19 at 13:35

gsamaras

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you means this as const char* str ? I can't get why this happens ? – ahmed roshdy May 12 '19 at 13:46
2

Because it stored in read-only memory @ahmedroshdy. Check [this](https://stackoverflow.com/questions/5464183/modifying-string-literal) for more. – gsamaras May 12 '19 at 13:46
1

@xing it is [platform-specific](https://stackoverflow.com/questions/2589949/string-literals-where-do-they-go) where string literals are stored. – gsamaras May 12 '19 at 13:51
@gsamaras it is compiler dependent or what ? that means this value allocated in memory after linkage (linker process) I mean not in compile time right ? but this why happen ? – ahmed roshdy May 12 '19 at 13:57
1

Read the link @ahmedroshdy I answered with to xing. Because this is how the language is designed. Hope that helps! :) – gsamaras May 12 '19 at 13:58
@gsamaras yes it is helpful also [This](https://stackoverflow.com/questions/5464183/modifying-string-literal) , sorry for asking more but I like to deep understand this. anyway to enclose this we can say that any pointer to character for any platform it must be in data segment and this is how the language designed right ?.. thanks – ahmed roshdy May 12 '19 at 14:07
1

Any string literal is designed by the language to be not modifiable @ahmedroshdy. – gsamaras May 12 '19 at 14:17
1

Any string literal is designed by the language to be not modifiable @ahmedroshdy. BTW, it would be nice to accept an answer from the two posted here, since both are correct. – gsamaras May 12 '19 at 14:25

score 1 · Answer 2 · edited Jun 20 '20 at 09:12

1

To be more precise:

char *str = "AHMEDROSHDY";

'str` is a pointer to the string literal. String literals in C are not modifacable

in the C standard:

The behavior is undefined in the following circumstances: ...

The program attempts to modify a string literal

edited Jun 20 '20 at 09:12

Community

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answered May 12 '19 at 14:17

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Is it applicable to pass pointer to character (after assigning string to it) to method and modify this string after without using array and why not?

2 Answers2