long-long | 易学教程

How do I check if A+B exceed long long? (both A and B is long long) [duplicate]

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问题 This question already has an answer here: How do I detect unsigned integer multiply overflow? 31 answers I have two numbers: A and B . I need to calculate A+B somewhere in my code. Both A and B are long long , and they can be positive or negative . My code runs wrong, and I suspect the problem happens when calculating A+B . I simply want to check if A+B exceed long long range. So, any method is acceptable, as I only use it for debug. 回答1: Overflow is possible only when both numbers have the

What kind of data type is “long long”?

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问题 I don't know this type. Is that the biggest one from all? I think it is an integer type, right? Or is it a floating point thing? Bigger than double? 回答1: According to C99 standard, long long is an integer type which is at least 64-bit wide. There are two integer 64-bit types specified: long long int and unsigned long long int So, yes, this is the biggest integer type specified by C language standard (C99 version). There is also long double type specified by C99. It's an extended precision

sprintf for unsigned _int64

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问题 I am having following code. output of second %d in sprintf is always shown as zero. I think i am specifying wrong specifiers. Can any one help me in getting write string with right values. And this has to achieved in posix standard. Thanks for inputs void main() { unsigned _int64 dbFileSize = 99; unsigned _int64 fileSize = 100; char buf[128]; memset(buf, 0x00, 128); sprintf(buf, "\nOD DB File Size = %d bytes \t XML file size = %d bytes", fileSize, dbFileSize); printf("The string is %s ", buf)

How do you format an unsigned long long int using printf?

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#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; } Output: My number is 8 bytes wide and its value is 285212672l. A normal number is 0. I assume this unexpected result is from printing the unsigned long long int . How do you printf() an unsigned long long int ? John Downey Use the ll (el-el) long-long modifier with the u (unsigned) conversion. (Works in windows, GNU). printf("%llu", 285212672); You may want

How do you format an unsigned long long int using printf?

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问题 #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; } Output: My number is 8 bytes wide and its value is 285212672l. A normal number is 0. I assume this unexpected result is from printing the unsigned long long int . How do you printf() an unsigned long long int ? 回答1: Use the ll (el-el) long-long modifier