String error output

Question

I got a code. It should give me an output that will erase the middle character between 'z' and 'p'. for example: zipZap("zipXzap"): expected [zpXzp] but found [z pXz p]

std::string zipZap(const std::string& str){
    string a = str;  
    string b = "";
    size_t len = str.length();
    for (size_t i = 0; i < len; i++){
        if (str[i] == 'z')
            if (str[i+2] == 'p')
                a[i+1] = ' ';
    }
    return a;
}

When i replaced a[i+1] = ''; it gave me an error.

It did what you wanted to do. Remove characters between z and p. — Augustus Francis, Feb 19 '14 at 18:22
Well, it doesn't remove any characters. It replaces them with spaces. — Lightness Races in Orbit, Feb 19 '14 at 18:22
Your code has `a[i+1] = ' '`, that puts *spaces* there, not removes the existing character. — crashmstr, Feb 19 '14 at 18:23
What string do you pass it, and what do you want it to return? — Beta, Feb 19 '14 at 18:33
for example, i pass [zipXzap] , it should return [zpXzp] with no space. But i give me a space between. — user3326237, Feb 19 '14 at 18:35
possible duplicate of [C++ Remove punctuation from String](http://stackoverflow.com/questions/19138983/c-remove-punctuation-from-string) and [Remove character from array where spaces and punctuation marks are found](http://stackoverflow.com/questions/21763038/remove-character-from-array-where-spaces-and-punctuation-marks-are-found/21763466#21763466) — πάντα ῥεῖ, Feb 19 '14 at 19:06

score 0 · Answer 1 · answered Feb 19 '14 at 18:35

You are not removing the chars, you are replacing them with ' '.

There are many ways to do this. One simple way is to build a new string, only adding chars when the proper conditions are met:

std::string zipZap(const std::string& str)
{
    string a;  
    size_t len = str.length();
    for (size_t i = 0; i < len; i++) {
        // Always add first and last chars. As well as ones not between 'z' and 'p'
        if (i == 0 || i == len-1 || (str[i-1] != 'z' && str[i+1] != 'p')) {
            a += str[i];
        }
    }
    return a;
}

score 0 · Answer 2 · answered Feb 19 '14 at 18:37

Use string.erase() :

std::string zipZap(const std::string& str){
    std::string a = str;
    std::string b = "";
    size_t len = str.length();
    for (size_t i = 0; i < len; i++){
        if (a[i] == 'z')
            if (a[i+2] == 'p')
                a.erase(i+1,1);
    }
    return a;
}

score 0 · Answer 3 · answered Feb 19 '14 at 18:41

You're completely right that you cant replace an element of the string with ''. A string is an array of chars, and '' is not a char at all. It is nothing. If we look at the cplusplus page for a string

http://www.cplusplus.com/reference/string/string/

We see that we can use erase(iterator p) to "Erase characters from string (public member function)"

So if we change:

for (size_t i = 0; i < len; i++){
    if (str[i] == 'z')
        if (str[i+2] == 'p')
            a.erase(a.begin() + i + 1);

We're closer now, but we can see that len is no longer the same as str.length(). the length of a is now actually 1 char shorter than len. To remedy this however we can simply add:

for (size_t i = 0; i < len; i++){
    if (str[i] == 'z')
        if (str[i+2] == 'p')
            a.erase(a.begin() + i + 1);
            len -= 1;

Hope that helps

score 0 · Answer 4 · answered Feb 19 '14 at 18:44

0

If you #include <regex>, you can do a regular expression replacement.

std::string zipZap(const std::string& str){
    regex exp("z.p");
    string a = str;
    a = regex_replace(a, exp "zp");
    return a;
}

answered Feb 19 '14 at 18:44

McLovin

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String error output

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