如何使用printf格式化unsigned long long int?

#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 intapi


#1樓

使用VS2005將其編譯爲x64: ide

%llu運做良好。 spa


#2樓

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

#3樓

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);

#4樓

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Lllq it

和windows下我相信這是llio


#5樓

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。

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