Replacing character in string in cpp using pointer

Question

I am very new new to cpp and trying to replace second occurrence of '*' with '!' char in a given string using following method.

#include <iostream>
#include <string.h>

using namespace std;

void replaceChar(char **inp){
    char *tmp = *inp;
    const char *c = "*";
    char *cmark = strstr(tmp,c);
    cout<< *cmark;
    if(cmark != NULL && strlen(cmark) > 1){
        cmark++;
        if(strstr(cmark,c)){
            int len = strlen(cmark);
            cout<<"len"<<len;
            for(int i=0;i<len;i++){
                if(cmark[i] == '*'){
                    cout<<"i.."<<i;

                    cmark[i] = '!';//error point
                }
            }
        }
    }

}


int main() {

    char * val = "this is string*replace next * with ! and print";
    replaceChar(&val);
    cout<<"val is "<< val;
    return 0;
}

I am getting run time error on error point line.If I comment out this line I am getting correct index of '*' to be replaced. Is it possible to replace '*' with '!' using cmark[i] = '!'?

Perhaps not trying to modify a read-only string literal may help. char val[] = .... And I see little sense in passing val by address to this function, btw, a problem you will need to fix if val is an array-type rather than the pointer it currently is. — WhozCraig
– WhozCraig, Commented Jul 7, 2016 at 7:09
It means if val is pointer type then it will be read only.Also replacing char by passing address is not possible? — user3512178
– user3512178, Commented Jul 7, 2016 at 7:21
The type isn't the problem; it is what it points to and what you're trying to do to that data where the wheels fall off the wagon. Your code declares a pointer that points to a read-only literal. The line you marked tries to write to that memory, and thus your program (fortunately) crashed. — WhozCraig
– WhozCraig, Commented Jul 7, 2016 at 7:24
Finally found the question that describes the root of your crash: See here. — WhozCraig
– WhozCraig, Commented Jul 7, 2016 at 7:30
Other than some stream helpers to make output easy, this is really a C question. — Donnie
– Donnie, Commented Jul 8, 2016 at 5:21

Community · Accepted Answer · 2017-05-23 11:58:56Z

1

Check this difference between char s[] and char *s in C

#include <iostream>
#include <string.h>

using namespace std;

void replaceChar(char *inp){
    char *tmp = inp;
    const char *c = "*";
    char *cmark = strstr(tmp,c);
    cout<< *cmark;
    if(cmark != NULL && strlen(cmark) > 1){
        cmark++;
        if(strstr(cmark,c)){
            int len = strlen(cmark);
            cout<<"len"<<len;
            for(int i=0;i<len;i++){
                if(cmark[i] == '*'){
                    cout<<"i.."<<i;

                    cmark[i] = '!';
                }
            }
        }
    }

}


int main() {

    char val[] = "this is string*replace next * with ! and print";
    replaceChar(val);
    cout<<"val is "<< val;
    return 0;
}

edited May 23, 2017 at 11:58

CommunityBot

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answered Jul 8, 2016 at 5:06

kid

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Comments

Mikhail · Accepted Answer · 2016-07-07 07:32:04Z

0

There is no need to pass pointer to pointer in method. Instead you can just pass original pointer to string. You can do it in much simpler way.

void replaceChar(char *inp){
    int i;
    int second = 0;
    /* Strings in C\C++ is null-terminated so we use it to determine 
    end of string */
    for (i = 0; inp[i] != '\0'; ++i) {
        if (inp[i] == '*') {
            /* Use flag to determine second occurrence of * */
            if (!second) {
                second = 1;
            } else {
                inp[i] = '!';
                break;
            }
        }
    }
}

edited Jul 7, 2016 at 7:32

answered Jul 7, 2016 at 7:27

Mikhail

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1 Comment

Donnie Over a year ago

Although correct, this does not address the root of the problem and will also crash.

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