The following fails with the error prog.cpp:5:13: error: invalid conversion from ‘char’ to ‘const char*’
int main()
{
char d = 'd';
std::string y("Hello worl");
y.append(d); // Line 5 - this fails
std::cout << y;
return 0;
}
I also tried, the following, which compiles but behaves randomly at runtime:
int main()
{
char d[1] = { 'd' };
std::string y("Hello worl");
y.append(d);
std::cout << y;
return 0;
}
Sorry for this dumb question, but I've searched around google, what I could see are just "char array to char ptr", "char ptr to char array", etc.
int main()
{
char d = 'd';
std::string y("Hello worl");
y += d;
y.push_back(d);
y.append(1, d); //appending the character 1 time
y.insert(y.end(), 1, d); //appending the character 1 time
y.resize(y.size()+1, d); //appending the character 1 time
y += std::string(1, d); //appending the character 1 time
}
Note that in all of these examples you could have used a character literal directly: y += 'd';
.
Your second example almost would have worked, for unrelated reasons. char d[1] = { 'd'};
didn't work, but char d[2] = { 'd'};
(note the array is size two) would have been worked roughly the same as const char* d = "d";
, and a string literal can be appended: y.append(d);
.
I found a simple way...
I needed to tack a char
on to a string that was being built on the fly. I needed a char list;
because I was giving the user a choice and using that choice in a switch()
statement.
I simply added another std::string Slist;
and set the new string equal to the character, "list" - a, b, c or whatever the end user chooses like this:
char list;
std::string cmd, state[], Slist;
Slist = list; //set this string to the chosen char;
cmd = Slist + state[x] + "whatever";
system(cmd.c_str());
Complexity may be cool but simplicity is cooler. IMHO
str.append(10u,'d'); //appends character d 10 times
Notice I have written 10u and not 10 for the number of times I'd like to append the character; replace 10 with whatever number.
To add a char to a std::string var using the append method, you need to use this overload:
std::string::append(size_type _Count, char _Ch)
Edit : Your're right I misunderstood the size_type parameter, displayed in the context help. This is the number of chars to add. So the correct call is
s.append(1, d);
not
s.append(sizeof(char), d);
Or the simpliest way :
s += d;
Try using the d as pointer y.append(*d)
I test the several propositions by running them into a large loop. I used microsoft visual studio 2015 as compiler and my processor is an i7, 8Hz, 2GHz.
long start = clock();
int a = 0;
//100000000
std::string ret;
for (int i = 0; i < 60000000; i++)
{
ret.append(1, ' ');
//ret += ' ';
//ret.push_back(' ');
//ret.insert(ret.end(), 1, ' ');
//ret.resize(ret.size() + 1, ' ');
}
long stop = clock();
long test = stop - start;
return 0;
According to this test, results are :
operation time(ms) note
------------------------------------------------------------------------
append 66015
+= 67328 1.02 time slower than 'append'
resize 83867 1.27 time slower than 'append'
push_back & insert 90000 more than 1.36 time slower than 'append'
Conclusion
+=
seems more understandable, but if you mind about speed, use append
the problem with:
std::string y("Hello worl");
y.push_back('d')
std::cout << y;
is that you have to have the 'd' as opposed to using a name of a char, like char d = 'd'; Or am I wrong?
If you are using the push_back there is no call for the string constructor. Otherwise it will create a string object via casting, then it will add the character in this string to the other string. Too much trouble for a tiny character ;)
Also adding insert option, as not mentioned yet.
std::string str("Hello World");
char ch;
str.push_back(ch); //ch is the character to be added
OR
str.append(sizeof(ch),ch);
OR
str.insert(str.length(),sizeof(ch),ch) //not mentioned above
In addition to the others mentioned, one of the string constructors take a char and the number of repetitions for that char. So you can use that to append a single char.
std::string s = "hell";
s += std::string(1, 'o');
Use push_back()
:
std::string y("Hello worl");
y.push_back('d')
std::cout << y;
Source: Stackoverflow.com