这怎么解决 warning C4307: “*”: 整型常量溢出

[河东的河西]

DWORD lAdd;
for(lAdd = 4*1024*1024;lAdd < 2*1024*1024*1024 ;lAdd += (4*1024))
{
Ret = CompareMemony((void *)lAdd,iVal);
if(Ret == FALSE) lAdd += (1*1024*1024)-(4*1024);
}
这个 CompareMemony函数读取进程数据用的
求解啊

[微型蚂蚁]

unsigned long long

[zhuben775]

一般来说，程序员会尽量简化乘法运算，像你上面的算法，1024完全就可以提出来，运用数学来避免溢出。 实在就是无法简化，就用ULONGLONG类型吧，如果ULONGLONG类型都无法解决，只能用大数运算了，那个就很费事了

[赵4老师]

__int64 lAdd;
for(lAdd = 4i64*1024i64*1024i64;lAdd < 2i64*1024i64*1024i64*1024i64 ;lAdd += (4i64*1024i64))
{

if( lAdd<(__int64)iVal) lAdd += (1i64*1024i64*1024i64)-(4i64*1024i64);
}

[赵4老师]

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

[阿呆_]

溢出在 2*1024*1024*1024，这个超过32位的signed int范围了（尽管还在32位unsigned int范围内）,改成1024u 应该就可以了。

[brk1985]

DWORD lAdd;是4个字节，溢出了，改成8字节的long long或unsigned long long就行了

[跑得快呢]

DWORD64 lAdd;
 for(lAdd = static_cast<DWORD64>(4)*1024*1024;lAdd < static_cast<DWORD64>(2)*1024*1024*1024 ;lAdd += (4*1024))
 {
 Ret = CompareMemony((void *)lAdd,iVal);
 if(Ret == FALSE) lAdd += (1*1024*1024)-(4*1024);
 }

做乘法时遇到整型常量溢出,只需要把第一个乘数强制转换为DWORD64