#include <stdio.h> int main() { unsigned long long int num = 285212672; //FYI: fits in 29 bits int normalInt = 5; printf("My number is %d bytes wide and its value is %ul. A normal number is %d.\n", sizeof(num), num, normalInt); return 0; }
輸出: windows
My number is 8 bytes wide and its value is 285212672l. A normal number is 0.
我認爲這種意外的結果是因爲打印unsigned long long int
。 你如何printf()
一個unsigned long long int
? api
使用VS2005將其編譯爲x64: ide
%llu運做良好。 spa
對於使用MSVS的很長一段時間(或__int64),應該使用%I64d: code
__int64 a; time_t b; ... fprintf(outFile,"%I64d,%I64d\n",a,b); //I is capital i
將ll(el-el)long-long修飾符與u(無符號)轉換一塊兒使用。 (在Windows,GNU中運行)。 orm
printf("%llu", 285212672);
非標準的東西老是很奇怪:) 編譯器
對於GNU下的long long部分,它是L
, ll
或q
it
和windows下我相信這是ll
只 io
好吧,一種方法是使用VS2008將其編譯爲x64 編譯
這將按您指望的那樣運行:
int normalInt = 5; unsigned long long int num=285212672; printf( "My number is %d bytes wide and its value is %ul. A normal number is %d \n", sizeof(num), num, normalInt);
對於32位代碼,咱們須要使用正確的__int64格式說明符%I64u。 就這樣。
int normalInt = 5; unsigned __int64 num=285212672; printf( "My number is %d bytes wide and its value is %I64u. A normal number is %d", sizeof(num), num, normalInt);
此代碼適用於32位和64位VS編譯器。