Use the .isdigit() method: >>> ‘123’.isdigit() True >>> ‘1a23′.isdigit() False Quoting the documentation: Return true if all characters in the string are digits and there is at least one character, false otherwise. For unicode strings or Python 3 strings, you’ll need to use a more precise definition and use the unicode.isdecimal() / str.isdecimal() instead; not … Read more
We have got to love list comprehensions. [dict([a, int(x)] for a, x in b.items()) for b in list] (Remark: for Python 2-only code, you may use “iteritems” instead of “items”)
If input value can be in numeric form other than integer , check by if (x == (int)x) { // Number is integer } If string value is being passed , use Integer.parseInt(string_var). Please ensure error handling using try catch in case conversion fails.
When writing in C or C++, every datatype is architecture and compiler specific. On one system int is 32, but you can find ones where it is 16 or 64; it’s not defined, so it’s up to compiler. As for long and int, it comes from times, where standard integer was 16bit, where long was … Read more
Now in c++11 we have #include <string> string s = std::to_string(123); Link to reference: http://en.cppreference.com/w/cpp/string/basic_string/to_string
You just need to cast at least one of the operands to a float: float z = (float) x / y; or float z = x / (float) y; or (unnecessary) float z = (float) x / (float) y;
Here is what the program does at each step: 1 * 10 = 10 10 * 11 = 110 110 * 12 = 1320 1320 * 13 = 17160 17160 * 14 = 240240 240240 * 15 = 3603600 3603600 * 16 = 57657600 57657600 * 17 = 980179200 980179200 * 18 = 463356416 463356416 … Read more
int(x) Conversion to integer will truncate (towards 0.0), like math.trunc. For non-negative numbers, this is downward. If your number can be negative, this will round the magnitude downward, unlike math.floor which rounds towards -Infinity, making a lower value. (Less positive or more negative). Python integers are arbitrary precision, so even very large floats can be … Read more
Yes, it is possible. For example, for gcc you can use __builtin_unreachable to tell the compiler about impossible conditions, like so: if (value < 0 || value > 36) __builtin_unreachable(); We can wrap the condition above in a macro: #define assume(cond) do { if (!(cond)) __builtin_unreachable(); } while (0) And use it like so: assume(x … Read more