programming question

[MisterWhite]

So there is a task:

Given a string, return a string where every appearance of the lowercase word "is" has been replaced with "is not".

notReplace("is test") → "is not test"
notReplace("is-is") → "is not-is not"
notReplace("This is right") → "This is not right"

 

i plan on doing this with c++;

i dont need the code of how it's done, just the main concept.

[ALwin]

Use regular expressions to check for the "is" and do a replacement of the string.

[Sauron]

Use regular expressions to check for the "is" and do a replacement of the string.

 

this is an option. Another option is to stuff every word in a list and print the list recursively. In the loop, check if the word is "is", in which case print "not" right after.

[MisterWhite]

Use regular expressions to check for the "is" and do a replacement of the string.

    string start = "There is an animal named is";    int countLetters = 0;    for(int i=0; i<start.length(); i++)    {        countLetters++;//used if == 2        if( start[i]==' ' || start[i+1]=='\n')// the part with \n does not work        {                if(start[i-3]==' ' && start[i-(countLetters-1)]=='i' && start[i-(countLetters-2)]=='s')                    // 1st condition is to check if after "is" (2 chars) goes ' ' 3rd char                    cout<<"is not ";            countLetters=0;        }    }

so i've made this brute code . though now it only prints "is not " if "is" is found (dont know how to include it in the main string). and it does not print if "is" is at the end.

[Nineshadow]

@MisterWhite

 

Here is my function.

It's pretty simple, it works with usual strings (array of characters) .

 

It modifies the actual string itself , rather than just printing the wanted result. I think it's nicer this way .

It takes as input a pointer to a character, basically a character array.

I think it's pretty easy to understand how it works.

Iterate through the array until you reach its end or you find an 'i' . If the next character after that one is 's' , that means we have found an 'is' . We shift the array to the right by 4 starting from the next position , then complete the array with our " not" string.

 

Tested it and it works.

void replaceNot(char *str){    int len = strlen(str);    for(int i = 0 ; i < len; i++)    {        if(str[i] == 'i' && str[i+1] == 's')        {            len+=4;            for(int j = len; j > i+5; j--)                str[j] = str[j-4];            str[i+2] = ' ';            str[i+3] = 'n';            str[i+4] = 'o';            str[i+5] = 't';        }    }}

You use it something like this :

int main(){    char c[100] = "This is right";    replaceNot(c);    cout << c;    return 0;}

[fizzlesticks]

It's pretty simple, it works with usual strings (array of characters) .

Or like this with real strings since it's C++.  It could be made more efficient if you have specific use cases in mind but here's a generic function.

std::string replace(std::string str, std::string sub, std::string repl){    size_t index = 0;    size_t subLen = sub.length();    size_t replLen = repl.length();    while((index = str.find(sub, index)) != std::string::npos)    {        str.replace(index, subLen, repl);        index += replLen;    }    return str;}int main(){    std::string s = replace("is asd is gneirng is not", "is", "is not");    return 0;} 

[Nineshadow]

 

Or like this with real strings since it's C++.  It could be made more efficient if you have specific use cases in mind but here's a generic function.

 

Ehehe , I'm working on a generic function for it as well. Just with c strings...it's a little more challenging

[Nineshadow]

 

Or like this with real strings since it's C++.  It could be made more efficient if you have specific use cases in mind but here's a generic function.

 

Bam, and I delivered!

void replace(char *dest , char *find , char *str){    int destlen = strlen(dest);    int strlenn = strlen(str);    int findlen = strlen(find);    bool OK;    for(int i = 0 ; i < destlen; i++)    {        OK = false;        if(dest[i] == find[0])        {            OK = true;            for(int j = 1; j < findlen ; j++)                if( dest[i+j] != find[j]) OK = false;        }        if(OK)        {            int delta = strlenn-findlen;            destlen+=delta;            if(delta >= 0)                for(int j = destlen; j >= i + strlenn; j--)                    dest[j] = dest[j-delta];            else            {                for(int j = i + strlenn; j < destlen; j++)                    dest[j] = dest[j-delta];                dest[destlen] = '\0';            }            for(int j = 0 ; j < strlenn; j++)                dest[i+j] = str[j];            i+=strlenn;        }    }}

It's generic. Works even if the string you are replacing with is shorter .