I want to pass string to ac 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
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, ie char str[] = "AHMEDROSHDY";
.
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
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.