Hi,
If I have a string like this:
char buff[10];
buff[0] ='h';
buff[1] ='e';
buff[2] ='l';
buff[3] ='l';
buff[4] ='o';
string s(buff);
How can i replace 'll' with 'abc'
and what if I replace 'll' with 'a', will the 'o' move up by 1 index?
Thank you. 7 7957 si************* **@gmail.com skrev:
Hi,
If I have a string like this:
char buff[10];
buff[0] ='h';
buff[1] ='e';
buff[2] ='l';
buff[3] ='l';
buff[4] ='o';
buff[5] = 0;
string s(buff);
How can i replace 'll' with 'abc'
and what if I replace 'll' with 'a', will the 'o' move up by 1 index?
replace(s, "ll", "abc");
void replace(std::st ring &target, std::string &that, std::string &with) {
std::string::si ze_type where = target.find(tha t);
if(where != std::string::np os) {
target.replace( target.begin() + where,
target.begin() + where + that.size(),
with.begin(),
with.end());
}
}
--
OU
On May 30, 10:04 am, "silverburgh.me ...@gmail.com"
<silverburgh.me ...@gmail.comwr ote:
Hi,
If I have a string like this:
char buff[10];
buff[0] ='h';
buff[1] ='e';
buff[2] ='l';
buff[3] ='l';
buff[4] ='o';
You need to insert a null character at postion 5.
string s(buff);
How can i replace 'll' with 'abc'
s.replace(2,2," abc");
and what if I replace 'll' with 'a', will the 'o' move up by 1 index?
yes
>
Thank you.
si************* **@gmail.com wrote:
Hi,
If I have a string like this:
char buff[10];
buff[0] ='h';
buff[1] ='e';
buff[2] ='l';
buff[3] ='l';
buff[4] ='o';
string s(buff);
How can i replace 'll' with 'abc'
and what if I replace 'll' with 'a', will the 'o' move up by 1 index?
Thank you.
Here are all the string class member functions: http://cppreference.com/cppstring/index.html
On Wed, 30 May 2007 19:30:52 +0200, Obnoxious User wrote:
[...]
>void replace(std::st ring &target, std::string &that, std::string &with) {
const correctness?
--
Roland Pibinger
"The best software is simple, elegant, and full of drama" - Grady Booch
Roland Pibinger skrev:
On Wed, 30 May 2007 19:30:52 +0200, Obnoxious User wrote:
[...]
>void replace(std::st ring &target, std::string &that, std::string &with) {
const correctness?
void replace(std::st ring & target,
std::string const & that,
std::string const & with);
--
OU
<si************ ***@gmail.comwr ote in message news:11******** *************@u 30g2000hsc.goog legroups.com...
Hi,
If I have a string like this:
char buff[10];
buff[0] ='h';
buff[1] ='e';
buff[2] ='l';
buff[3] ='l';
buff[4] ='o';
Triply wrong:
1. Firstly, null-terminated strings need to be, well, null-terminated.
So you'd need to set:
buff[5] = '\0';
2. Why not just initialize the buffer to the string?
char buff[10] = "hello";
3. Why use a char[] buffer at all? Just do this:
std::string MyString ("hello");
string s(buff);
std::string
How can i replace 'll' with 'abc'
Like so (tested, working program; compile this for demo):
#include <iostream>
#include <string>
int main (void)
{
std::string S ("hello");
S.replace(S.fin d("ll"), 2, "abc"); // replace 2 chars at "ll" by "abc"
std::cout << S << std::endl;
return 0;
}
and what if I replace 'll' with 'a', will the 'o' move up by 1 index?
It decrements by one, from 4 to 3.
Thank you.
You're welcome.
--
Cheers,
Robbie Hatley
lone wolf aatt well dott com
triple-dubya dott Tustin Free Zone dott org
"Obnoxious User" <OU@127.0.0.1wr ote in message
news:46******** *************** @alt.teranews.c om...
replace(s, "ll", "abc");
void replace(std::st ring &target, std::string &that, std::string &with) {
std::string::si ze_type where = target.find(tha t);
if(where != std::string::np os) {
target.replace( target.begin() + where,
target.begin() + where + that.size(),
with.begin(),
with.end());
}
}
Ewww. I could replace all of the above with just this:
s.replace(s.fin d("ll"), 2, "abc");
or more generically:
s.replace(s.fin d(target), target_size, replacement);
Far more convenient, and far less prone to error. "find" and "replace" are
already built into the std::string class, so might as well use them.
--
Cheers,
Robbie Hatley
lone wolf aatt well dott com
triple-dubya dott Tustin Free Zone dott org This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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