well the subject tells the story..
should be simple to do.. but my brain is drawing a blank
i have ...
int i;
string ret="A";
...
...
...
i need to append the value of i to the ret string...
the int is just a counter in a loop, the value of the string needs to end up being somthing like.. A1 or A2 ..
danka
Don't forget to include <string>!Code:string ret("A"); for(int i=50; i<=57; i++) // 50 is 1, 57 is 9 { ret[1] = i; }
thanks for the reply.. but i got it.. i wasnt thinking!!!
for some reason i didnt think sprintf would print 2 vars to a string... so i fixed it like..
ret is now a char * ..
sprintf(ret, "%c%i", theAlphabet, i);
its alive!!!!!! hahahahahaha....
You might try:
int i = 1;
string ret="A";
char temp[100];
do{
sprintf( temp, "%i", i++ );
ret += temp;
}while(/*condition*/);
Actually, though, this would result in:
"A1", "A12", then"A123", etc...
Well you could do:
int i = 1;
string prep = "A";
char temp[100];
do{
string ret = prep;
sprintf( temp, "%i", i++);
ret += temp;
}while(/*condition*/);
Code:#include <cmath> #include <complex> bool euler_flip(bool value) { return std::pow ( std::complex<float>(std::exp(1.0)), std::complex<float>(0, 1) * std::complex<float>(std::atan(1.0) *(1 << (value + 2))) ).real() < 0; }
