Hi,
I have a following problem. I have a number in string, for example "123456" and I'd like to convert it to int. This would not be a problem if I knew that the input number will always fit into an 32 bit integer. So if it won't I should take only the first 31 bit of that number, show some error and the program should continue. I know that there is the atoi function which I think can handle the overflow but I need to write my own function and I don't know how.
Thx for any help.
You should use the atoi function in C++. Conversion are done easier with the use of a stringstream. By checking the stream after the conversion, we can see if the conversion has succeeded.
#include <sstream>
#include <string>
#include <iostream>
#include <exception>
int strToInt(std::string in) {
int ret_val = 0;
std::stringstream sstr(in);
sstr >> ret_val;
if (sstr.fail()) { //Check if conversion worked
throw std::exception(std::string("Failed conversion for: " + in).c_str());
}
return ret_val;
}
int main() {
try {
int i = strToInt("1000000"); // Will fit
int j = strToInt("20000000000"); // Wont fit
} catch (std::exception &ex) {
std::cout << "ERROR: " << ex.what();
}
}You should use the atoi function in C++. Conversion are done easier with the use of a stringstream. By checking the stream after the conversion, we can see if the conversion has succeeded.
#include <sstream> #include <string> #include <iostream> #include <exception> int strToInt(std::string in) { int ret_val = 0; std::stringstream sstr(in); sstr >> ret_val; if (sstr.fail()) { //Check if conversion worked throw std::exception(std::string("Failed conversion for: " + in).c_str()); } return ret_val; } int main() { try { int i = strToInt("1000000"); // Will fit int j = strToInt("20000000000"); // Wont fit } catch (std::exception &ex) { std::cout << "ERROR: " << ex.what(); } }
Thx for the reply, this really works but if it won't fit I need the first 31 bits of that number and the rest throw away. Any idea?
You know bitwise operators?
You know bitwise operators?
Yes I know them but still I don't know how it has to be done
I'm not going to give it away. :)
Let's start with a smaller number, like 7.
which number and which operator do you need to get the last bit (=1 because 7 = 0111 binary).
0111
xxxx
----- ??? operator
0001Choose from or, and, xor.
I'm not going to give it away. :)
Let's start with a smaller number, like 7.which number and which operator do you need to get the last bit (=1 because 7 = 0111 binary).
0111 xxxx ----- ??? operator 0001Choose from or, and, xor.
Ok 1001 and
Actually 0001 and. You put a 1 at each position you want the bit from. So in your case you need the last 31 bits which is 01111111111111111111111111111111 which is 2147483647 decimal. So if you do int last31 = youinput & 2147483647; you should be done.
Actually 0001 and. You put a 1 at each position you want the bit from. So in your case you need the last 31 bits which is 01111111111111111111111111111111 which is 2147483647 decimal. So if you do
int last31 = youinput & 2147483647;you should be done.
Thanks again,I get it but I still don't know how to get the bit array of the number which is stored in that string.
I don't understand. Do you want
1. The last 31 bits as a decimal number?
2. The last 31 bits as an array (bitset) of 1's and 0's?
I don't understand. Do you want
1. The last 31 bits as a decimal number?
2. The last 31 bits as an array (bitset) of 1's and 0's?
What I exactly need is: I will get an string which contains a really large number like "123456789123456789" which don't fit into an integer and I need the first 31 bits of this number(not string).
Yeah, that part I get.
But do you want the 31 bits as a number (like 7274749912) or as bits (like 11100011100101010010)
Yeah, that part I get.
But do you want the 31 bits as a number (like 7274749912) or as bits (like 11100011100101010010)
If I will have the decimal or binary I can simply convert it to another, so it doesn't matter I just need to know how to get that (especially in c++).
In that case I already told you the answer.
Integrated in the program:
#include <sstream>
#include <string>
#include <iostream>
#include <exception>
int strToInt(std::string in) {
unsigned long long ret_val = 0;
std::stringstream sstr(in);
sstr >> ret_val;
if (sstr.fail()) { //Check if conversion worked
throw std::exception(std::string("Failed conversion for: " + in).c_str());
}
return static_cast<int>(ret_val & 2147483647); //use bitwise & with the last 31 bits.
}
int main() {
try {
std::cout << strToInt("1234567890") << '\n';// show entire number, because it fits an int
std::cout << strToInt("12345678900")<< '\n';// Only last 31 bits because number is too big.
} catch (std::exception &ex) {
std::cout << "ERROR: " << ex.what();
}
}In that case I already told you the answer.
Integrated in the program:
#include <sstream> #include <string> #include <iostream> #include <exception> int strToInt(std::string in) { unsigned long long ret_val = 0; std::stringstream sstr(in); sstr >> ret_val; if (sstr.fail()) { //Check if conversion worked throw std::exception(std::string("Failed conversion for: " + in).c_str()); } return static_cast<int>(ret_val & 2147483647); //use bitwise & with the last 31 bits. } int main() { try { std::cout << strToInt("1234567890") << '\n';// show entire number, because it fits an int std::cout << strToInt("12345678900")<< '\n';// Only last 31 bits because number is too big. } catch (std::exception &ex) { std::cout << "ERROR: " << ex.what(); } }
Alright this works good but I have two more question. Why does it fail on number bigger than 123456789012345678901 (10 times bigger). And I wanted the first 31 bits not the last.
These are the first 31 bits. The first bit is the one on the right (also known as lsb)
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