user-0 提问: uint64_t的运算出错问题

回复:
user: user-1
content: y = 3000ui64 * 24000000ui64 / 1000ui64; C/C++语言整数常量语法（参考3楼）不过关。

user: user-0
content: 楼上都是专家啊，学习了，谢谢！

user: user-2
content: y = (uint64_t)3000 * (uint64_t)24000000 / 1000;  C语言基础不过关。

user: user-1
content: 3000和3000i64和3000ui64不是一回事！

user: user-1
content: 常量也有类型！ C++ Integer Constants Integer constants are constant data elements that have no fractional parts or exponents. They always begin with a digit. You can specify integer constants in decimal, octal, or hexadecimal form. They can specify signed or unsigned types and long or short types.  Syntax  integer-constant :  decimal-constant integer-suffixopt octal-constant integer-suffixopt hexadecimal-constant integer-suffixopt 'c-char-sequence'  decimal-constant :  nonzero-digit decimal-constant digit  octal-constant :  0 octal-constant octal-digit  hexadecimal-constant :  0x hexadecimal-digit 0X hexadecimal-digit hexadecimal-constant hexadecimal-digit  nonzero-digit : one of  1 2 3 4 5 6 7 8 9  octal-digit : one of  0 1 2 3 4 5 6 7  hexadecimal-digit : one of  0 1 2 3 4 5 6 7 8 9 a b c d e f A B C D E F  integer-suffix :  unsigned-suffix long-suffixopt long-suffix unsigned-suffixopt  unsigned-suffix : one of  u U  long-suffix : one of  l L  64-bit integer-suffix :  i64  To specify integer constants using octal or hexadecimal notation, use a prefix that denotes the base. To specify an integer constant of a given integral type, use a suffix that denotes the type.  To specify a decimal constant, begin the specification with a nonzero digit. For example:  int i = 157;   // Decimal constant int j = 0198;  // Not a decimal number; erroneous octal constant int k = 0365;  // Leading zero specifies octal constant, not decimal  To specify an octal constant, begin the specification with 0, followed by a sequence of digits in the range 0 through 7. The digits 8 and 9 are errors in specifying an octal constant. For example:  int i = 0377;   // Octal constant int j = 0397;   // Error: 9 is not an octal digit  To specify a hexadecimal constant, begin the specification with 0x or 0X (the case of the “x” does not matter), followed by a sequence of digits in the range 0 through 9 and a (or A) through f (or F). Hexadecimal digits a (or A) through f (or F) represent values in the range 10 through 15. For example:  int i = 0x3fff;   // Hexadecimal constant int j = 0X3FFF;   // Equal to i  To specify an unsigned type, use either the u or U suffix. To specify a long type, use either the l or L suffix. For example:  unsigned uVal = 328u;             // Unsigned value long lVal = 0x7FFFFFL;            // Long value specified                                    //  as hex constant unsigned long ulVal = 0776745ul;  // Unsigned long value

user: user-0
content: 引用 1 楼 turingo 的回复:y = 3000 * 24000000 / 1000; 溢出了。  y的定义是uint64_t y; 64位的，怎么就溢出了？

user: user-3
content: y = 3000 * 24000000 / 1000; 溢出了。  引用 楼主 syrchina 的回复:各位大神请看下面的程序，我的问题在注释里  //编译环境：codeblocks+gcc  #include <stdio.h> #include <stdint.h> int Fun() { 	uint64_t y; 	uint32_t x1, x2;      y = 3000 * 24000000 / 1000; 	printf("0x%llX\n", y);//输出错误结果：0xFFFFFFFFFFF08554 ，Why??? int64位不能正确运算吗？  	x1 = (uint32_t)(y & 0xFFFFFFFF);     x2 = (uint32_t)(y >> 32);     printf("0x%X,0x%X \n", x1, x2);//输出错误结果：0xFFF08554, 0xFFFFFFFF 	     y = 3000 * (24000000 / 1000); 	printf("0x%llX\n", y);//输出正确结果： 0x44AA200 	x1 = (uint32_t)(y & 0xFFFFFFFF);     x2 = (uint32_t)(y >> 32);     printf("0x%X,0x%X \n", x1, x2);//正确结果 ：0x44AA200， 0x0  } int main() { 	Fun(); }