long-long

I'd like to convert a long long to a string in C. long long x = 999; I'd like to convert x to a string. How could I go about doing that? Thanks. long long x = 999; char str[256]; sprintf(str, "%lld", x); printf("%s\n", str); 来源： https://stackoverflow.com/questions/16095248/convert-long-long-to-string-in-c

Wouldn't it have made more sense to make long 64-bit and reserve long long until 128-bit numbers become a reality? Yes, it does make sense, but Microsoft had their own reasons for defining "long" as 32-bits. As far as I know, of all the mainstream systems right now, Windows is the only OS where "long" is 32-bits. On Unix and Linux, it's 64-bit. All compilers for Windows will compile "long" to 32-bits on Windows to maintain compatibility with Microsoft. For this reason, I avoid using "int" and "long". Occasionally I'll use "int" for error codes and booleans (in C), but I never use them for any

I have an open-source codebase that is written in both C and C++. I'm looking for an integer type that is guaranteed to be at least 64 bits wide, which can be reliably compiled on most OS X (Intel, 64-bit) and Linux boxes with open-source C and C++ compilers, without too much extra work on the end user's part. Windows and 32-bit client support are not important at this time. I did some testing on OS X, and the latest GCC that ships with the developer tools does not support C+11 mode (and therefore does not seem to guarantee availability of long long ). Clang does not support this, either,

Given the following snippet: #include <stdio.h> typedef signed long long int64; typedef signed int int32; typedef signed char int8; int main() { printf("%i\n", sizeof(int8)); printf("%i\n", sizeof(int32)); printf("%i\n", sizeof(int64)); int8 a = 100; int8 b = 100; int32 c = a * b; printf("%i\n", c); int32 d = 1000000000; int32 e = 1000000000; int64 f = d * e; printf("%I64d\n", f); } The output with MinGW GCC 3.4.5 is (-O0): 1 4 8 10000 -1486618624 The first multiplication is casted to an int32 internally (according to the assembler output). The second multiplication is not casted. I'm not sure

This question already has an answer here: How do I detect unsigned integer multiply overflow? 31 answers I have two numbers: A and B . I need to calculate A+B somewhere in my code. Both A and B are long long , and they can be positive or negative . My code runs wrong, and I suspect the problem happens when calculating A+B . I simply want to check if A+B exceed long long range. So, any method is acceptable, as I only use it for debug. Daniel Fischer Overflow is possible only when both numbers have the same sign. If both are positive, then you have overflow if mathematically A + B > LLONG_MAX ,

I don't know this type. Is that the biggest one from all? I think it is an integer type, right? Or is it a floating point thing? Bigger than double? According to C99 standard, long long is an integer type which is at least 64-bit wide. There are two integer 64-bit types specified: long long int and unsigned long long int So, yes, this is the biggest integer type specified by C language standard (C99 version). There is also long double type specified by C99. It's an extended precision floating point numeric data type long for 80-bits on most popular x86-based platforms and implementations of C

I am having following code. output of second %d in sprintf is always shown as zero. I think i am specifying wrong specifiers. Can any one help me in getting write string with right values. And this has to achieved in posix standard. Thanks for inputs void main() { unsigned _int64 dbFileSize = 99; unsigned _int64 fileSize = 100; char buf[128]; memset(buf, 0x00, 128); sprintf(buf, "\nOD DB File Size = %d bytes \t XML file size = %d bytes", fileSize, dbFileSize); printf("The string is %s ", buf); } Output: The string is OD DB File Size = 100 bytes XML file size = 0 bytes DevSolar I don't know