#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;
}
输出:
My number is 8 bytes wide and its value is 285212672l. A normal number is 0.
我假设这个意外的结果来自于输出unsigned long long int。你如何打印()一个unsigned long long int?
当前回答
%d——>为int
%u——>为unsigned int
%ld—>为长int或长
%lu——>用于unsigned long int或long unsigned int或unsigned long
%lld—>表示long long int或long long
%llu——>用于unsigned long long int或unsigned long long
其他回答
您可能想尝试使用inttypes.h库，它可以为您提供诸如 Int32_t, int64_t, uint64_t等。 然后你可以使用它的宏，如:
#include <inttypes.h>
uint64_t x;
uint32_t y;
printf("x: %"PRIu64", y: %"PRIu32"\n", x, y);
这“保证”不会给您带来与long、unsigned long等相同的麻烦，因为您不必猜测每种数据类型中有多少位。
使用VS2005将其编译为x64:
%llu运行良好。
如何使用printf格式化unsigned long long int ?
由于C99在转换说明符o,u,x, x之前使用了“ll”(ell-ell)。
很多答案中除了10进制选项外，还有16进制和8进制选项:
选择包括
unsigned long long num = 285212672;
printf("Base 10: %llu\n", num);
num += 0xFFF; // For more interesting hex/octal output.
printf("Base 16: %llX\n", num); // Use uppercase A-F
printf("Base 16: %llx\n", num); // Use lowercase a-f
printf("Base 8: %llo\n", num);
puts("or 0x,0X prefix");
printf("Base 16: %#llX %#llX\n", num, 0ull); // When non-zero, print leading 0X
printf("Base 16: %#llx %#llx\n", num, 0ull); // When non-zero, print leading 0x
printf("Base 16: 0x%llX\n", num); // My hex fave: lower case prefix, with A-F
输出
Base 10: 285212672
Base 16: 11000FFF
Base 16: 11000fff
Base 8: 2100007777
or 0x,0X prefix
Base 16: 0X11000FFF 0
Base 16: 0x11000fff 0
Base 16: 0x11000FFF
%d——>为int
%u——>为unsigned int
%ld—>为长int或长
%lu——>用于unsigned long int或long unsigned int或unsigned long
%lld—>表示long long int或long long
%llu——>用于unsigned long long int或unsigned long long
对于使用MSVS的long long(或__int64)，您应该使用%I64d:
__int64 a;
time_t b;
...
fprintf(outFile,"%I64d,%I64d\n",a,b); //I is capital i
