Testing whether input is a valid int decimal numeral or is a decimal numeral in [-231, 231) is actually a bit complicated. The C standard does not provide a direct way to do this. What we can do is:
- Read characters and check to see whether they are in the expected form: spaces, an optional minus sign (hyphen), and digits. (Any non-digits after the digits will be allowed and ignored.)
- Try using
strtol to convert the numeral to a long. We use strtol because there is no C-standard library routine for converting to an int (or your fixed bounds using 231) that provides error indications.
- Compare the
long produced by strtol to the int bounds.
Example code for int bounds follows. If you want bounds of -2147483648 and 2147483647 instead, substitute those for INT_MIN and INT_MAX. To be completely safe, the code should actually use long long and strtoll, since the C standard does not require long to be able to represent −2147483648.
#include <ctype.h>
#include <errno.h>
#include <limits.h>
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
// Prepare a buffer.
size_t BufferSize = 100, BufferUsed = 0;
char *Buffer = malloc(BufferSize * sizeof *Buffer);
// Skip white space.
int c;
do
c = getchar();
while (isspace(c));
if (c == EOF)
{
printf("Input is not an int: EOF before \"-\" or digit seen.\n");
exit(EXIT_SUCCESS);
}
// Accept a hyphen as a minus sign.
if (c == '-')
{
Buffer[BufferUsed++] = c;
c = getchar();
}
// Accept digits.
while (isdigit(c))
{
Buffer[BufferUsed++] = c;
if (BufferSize <= BufferUsed)
{
BufferSize *= 2;
printf("Realloc: size = %zu, used = %zu.\n", BufferSize, BufferUsed);
char *NewBuffer = realloc(Buffer, BufferSize * sizeof *NewBuffer);
if (!NewBuffer)
{
fprintf(stderr, "Error, unable to allocate %zu bytes.\n",
BufferSize);
exit(EXIT_FAILURE);
}
Buffer = NewBuffer;
}
c = getchar();
}
// Ensure we saw at least one digit (input is not blank or just a hyphen).
if (BufferUsed == 0 || BufferUsed == 1 && Buffer[0] == '-')
{
printf("Input is not an int: No digits present.\n");
exit(EXIT_SUCCESS);
}
// Put back the unaccepted character, if any.
if (c != EOF)
ungetc(c, stdin);
// Terminate the string.
Buffer[BufferUsed] = 0;
// Attempt to convert the numeral to long.
char *End;
errno = 0;
long x = strtol(Buffer, &End, 10);
// Test whether strtol succeeded.
if (*End)
{
/* I do not expect this to occur since we already tested the input
characters.
*/
printf("Input is not an int: strtol rejected %c.\n", *End);
exit(EXIT_SUCCESS);
}
if (errno == ERANGE)
{
printf("Input is not an int: strtol reported out of range.\n");
exit(EXIT_SUCCESS);
}
if (x < INT_MIN || INT_MAX < x)
{
printf("Input is not an int: Value is outside bounds.\n");
exit(EXIT_SUCCESS);
}
printf("Input is an int, %ld.\n", x);
free(Buffer);
}
xof typeintcan't possibly fall outside the range representable inint, by definition.cin, cout,inheritance and overloading of functions. The C language doesn't.powfunction in C++ returns a floating point value. You may get inaccuracies do to conversion from floating point to integer. Best to use an integer exponentiation function.strtol()is a good starting point.pow(3, 31)is larger than(pow(2, 32) - 1).