floating-point-conversion

问题 This question already has an answer here : Closed 6 years ago . Possible Duplicate: Output 1000000 as 1,000,000 and so on I have a float variable in the format xxxxxxxx.xx (Eg. 11526.99). I'd like to print it as 11,562.99 with a comma. How can I insert a comma in C? 回答1: Try: #include <locale.h> #include <stdio.h> int main() { float f = 12345.67; // obtain the existing locale name for numbers char *oldLocale = setlocale(LC_NUMERIC, NULL); // inherit locale from environment setlocale(LC

问题 This question already has an answer here : Closed 6 years ago . Possible Duplicate: Output 1000000 as 1,000,000 and so on I have a float variable in the format xxxxxxxx.xx (Eg. 11526.99). I'd like to print it as 11,562.99 with a comma. How can I insert a comma in C? 回答1: Try: #include <locale.h> #include <stdio.h> int main() { float f = 12345.67; // obtain the existing locale name for numbers char *oldLocale = setlocale(LC_NUMERIC, NULL); // inherit locale from environment setlocale(LC

问题 I'm doing Learn Python the Hard Way exercise 35. Below is the original code, and we're asked to change it so it can accept numbers that don't have just 0 and 1 in them. def gold_room(): print "This room is full of gold. How much do you take?" next = raw_input("> ") if "0" in next or "1" in next: how_much = int(next) else: dead("Man, learn to type a number.") if how_much < 50: print "Nice, you're not greedy, you win!" exit(0) else: dead("You greedy bastard!") This is my solution, which runs

问题 I'm doing Learn Python the Hard Way exercise 35. Below is the original code, and we're asked to change it so it can accept numbers that don't have just 0 and 1 in them. def gold_room(): print "This room is full of gold. How much do you take?" next = raw_input("> ") if "0" in next or "1" in next: how_much = int(next) else: dead("Man, learn to type a number.") if how_much < 50: print "Nice, you're not greedy, you win!" exit(0) else: dead("You greedy bastard!") This is my solution, which runs

问题 int main() { float x=3.4e2; printf("%f",x); return 0; } Output: 340.000000 // It's ok. But if write x=3.1234e2 the output is 312.339996 and if x=3.12345678e2 the output is 312.345673 . Why are the outputs like these? I think if I write x=3.1234e2 the output should be 312.340000 , but the actual output is 312.339996 using GCC compiler. 回答1: Not all fractional numbers have an exact binary equivalent so it is rounded to the nearest value. Simplified example, if you have 3 bits for the fraction,

问题 I'm trying to re-construct a 32-bit floating point value from an eeprom. The 4 bytes in eeprom memory (0-4) are : B4 A2 91 4D and the PC (VS Studio) reconstructs it correctly as 3.054199 * 10^8 (the floating point value I know should be there) Now I'm moving this eeprom to be read from an 8-bit Arduino, so not sure if it's compiler/platform thing, but when I try reading the 4 bytes into a 32-bit dword, and then typecast it to a float, the value I get isn't even close. Assuming the conversion

问题 I'm trying to extract Zip Codes from an Excel Spreadsheet and load them into a list as Strings. import xlrd BIL = xlrd.open_workbook(r"C:\Temp\Stores.xls) Worksheet = BIL.sheet_by_name("Open_Locations") ZIPs = [] for record in Worksheet.col(17): if record.value == "Zip": pass else: ZIPs.append(record.value) Unfortunately, this Excel workbook is managed by someone else so I cannot simply go and convert the field holding zip codes in the excel spreadsheet to text to solve my problem. In

问题 I tried the equivalent of Michael Meadows EDIT 2, but in VB.NET and got a different result. (Specifically both the Double and Decimal results were 600000.0238418580.) I have determined the difference is with the compile-time accuracy of a float ( Single ) division stored into a float in C#, (which seems to be more equivalent to VB.NET's accuracy when storing into a Double ) and what happens (in both languages unsurprisingly) when you force the division to occur at runtime. So, THREE_FIFTHS

问题 I have written a program in C to convert a floating point number represented in binary ( 1101.11 ) into a decimal ( 13.75 ). However, I cannot seem to get the correct value out of the algorithm. What is the correct method for converting a binary floating point number into a decimal? I am using Dev CPP compiler (32 bit). The algorithm is defined below: void b2d(double p, double q ) { double rem, dec=0, main, f, i, t=0; /* integer part operation */ while ( p >= 1 ) { rem = (int)fmod(p, 10); p =

问题 Let's say I have two arithmetic types, an integer one, I , and a floating point one, F . I also assume that std::numeric_limits<I>::max() is smaller than std::numeric_limits<F>::max() . Now, let's say I have a positive integer value i . Because the representable range of F is larger than I , F(i) should always be defined behavior. However, if I have a floating point value f such that f == F(i) , is I(f) well defined? In other words, is I(F(i)) always defined behavior? Relevant section from