I need to get an argument and convert it to an int. Here is my code so far:
#include <iostream>
using namespace std;
int main(int argc,int argvx[]) {
int i=1;
int answer = 23;
int temp;
// decode arguments
if(argc < 2) {
printf("You must provide at least one argument\n");
开发者_运维百科 exit(0);
}
// Convert it to an int here
}
Since this answer was somehow accepted and thus will appear at the top, although it's not the best, I've improved it based on the other answers and the comments.
The C way; simplest, but will treat any invalid number as 0:
#include <cstdlib>
int x = atoi(argv[1]);
The C way with input checking:
#include <cstdlib>
errno = 0;
char *endptr;
long int x = strtol(argv[1], &endptr, 10);
if (endptr == argv[1]) {
std::cerr << "Invalid number: " << argv[1] << '\n';
} else if (*endptr) {
std::cerr << "Trailing characters after number: " << argv[1] << '\n';
} else if (errno == ERANGE) {
std::cerr << "Number out of range: " << argv[1] << '\n';
}
The C++ iostreams way with input checking:
#include <sstream>
std::istringstream ss(argv[1]);
int x;
if (!(ss >> x)) {
std::cerr << "Invalid number: " << argv[1] << '\n';
} else if (!ss.eof()) {
std::cerr << "Trailing characters after number: " << argv[1] << '\n';
}
Alternative C++ way since C++11:
#include <stdexcept>
#include <string>
std::string arg = argv[1];
try {
std::size_t pos;
int x = std::stoi(arg, &pos);
if (pos < arg.size()) {
std::cerr << "Trailing characters after number: " << arg << '\n';
}
} catch (std::invalid_argument const &ex) {
std::cerr << "Invalid number: " << arg << '\n';
} catch (std::out_of_range const &ex) {
std::cerr << "Number out of range: " << arg << '\n';
}
All four variants assume that argc >= 2
. All accept leading whitespace; check isspace(argv[1][0])
if you don't want that. All except atoi
reject trailing whitespace.
Note that your main
arguments are not correct. The standard form should be:
int main(int argc, char *argv[])
or equivalently:
int main(int argc, char **argv)
There are many ways to achieve the conversion. This is one approach:
#include <sstream>
int main(int argc, char *argv[])
{
if (argc >= 2)
{
std::istringstream iss( argv[1] );
int val;
if (iss >> val)
{
// Conversion successful
}
}
return 0;
}
std::stoi from string could also be used.
#include <string>
using namespace std;
int main (int argc, char** argv)
{
if (argc >= 2)
{
int val = stoi(argv[1]);
// ...
}
return 0;
}
As WhirlWind has pointed out, the recommendations to use atoi
aren't really very good. atoi
has no way to indicate an error, so you get the same return from atoi("0");
as you do from atoi("abc");
. The first is clearly meaningful, but the second is a clear error.
He also recommended strtol
, which is perfectly fine, if a little bit clumsy. Another possibility would be to use sscanf
, something like:
if (1==sscanf(argv[1], "%d", &temp))
// successful conversion
else
// couldn't convert input
note that strtol
does give slightly more detailed results though -- in particular, if you got an argument like 123abc
, the sscanf
call would simply say it had converted a number (123), whereas strtol
would not only tel you it had converted the number, but also a pointer to the a
(i.e., the beginning of the part it could not convert to a number).
Since you're using C++, you could also consider using boost::lexical_cast
. This is almost as simple to use as atoi
, but also provides (roughly) the same level of detail in reporting errors as strtol
. The biggest expense is that it can throw exceptions, so to use it your code has to be exception-safe. If you're writing C++, you should do that anyway, but it kind of forces the issue.
Take a look at strtol(), if you're using the C standard library.
The approach with istringstream can be improved in order to check that no other characters have been inserted after the expected argument:
#include <sstream>
int main(int argc, char *argv[])
{
if (argc >= 2)
{
std::istringstream iss( argv[1] );
int val;
if ((iss >> val) && iss.eof()) // Check eofbit
{
// Conversion successful
}
}
return 0;
}
Like that we can do....
int main(int argc, char *argv[]) {
int a, b, c;
*// Converting string type to integer type
// using function "atoi( argument)"*
a = atoi(argv[1]);
b = atoi(argv[2]);
c = atoi(argv[3]);
}
In my opinion this is the most practical way to pass your variables. Using std::stoi. It's good practice to use std:: and NOT to use namespace; this is why I can define a string with "string" as the variable name
#include <iostream>
#include <string>
int main(int argc, char **argv)
{
std::string string{argv[1]};
int integer1{std::stoi(argv[2])};
int integer2{std::stoi(argv[3])};
int sum{integer1 + integer2};
std::cout << string << " sum " << sum;
}
Example:
./out this 40 1 // input
this sum 41 // output
精彩评论