c++ - printf 和 scanf 中 %lli 和 %lld 之间的区别？

Question

这个问题已经有答案了:

What is the difference between conversion specifiers %i and %d in formatted IO functions (*printf / *scanf) (4 个回答)

已关闭 5 年前。

我和我的 friend 正在争论%lli和%lld 。我一般用%lli ，每次他都主张使用%lld在printf和scanf 。他声称%lli和%lld是不同的。

%lli 之间有什么区别吗？和%lld GNU GCC 编译器中的格式说明符，或者它们是否相同？

Answer 1

对于 printf 来说，它们是完全相同的。

http://en.cppreference.com/w/cpp/io/c/fprintf

d, i converts a signed integer into decimal representation [-]dddd.

Precision specifies the minimum number of digits to appear. The default precision is 1. If both the converted value and the precision are ​0​ the conversion results in no characters.

<小时/>

对于scanf，它们是不同的。以下是文档中的引用。

d matches a decimal integer. The format of the number is the same as expected by strtol() with the value 10 for the base argument

i matches an integer. The format of the number is the same as expected by strtol() with the value ​0​ for the base argument (base is determined by the first characters parsed)

对于i，如果您的号码以0开头，它将被解析为八进制。

http://en.cppreference.com/w/cpp/io/c/fscanf