我不知道为什么它发生！ 想知道原因。
{
int i=01;
printf("%d\n",i);
}
output: 1
但
{
int i=011;
printf("%d\n",i);
}
output: 9
没有任何人有答案吗？
Answer 1:
011 =八路，（1 * 8）+ 1 = 9 ........................
Answer 2:
011是一个八进制常数。 11 (b8) = 9 (b10)
C11（n1570），第6.4.4.1整型常量
一个八进制常数由前缀0可选地通过只7随后的数字0的序列组成。
Answer 3:
The numbers which are preceded by 0 is called octal numbers in c programming .
to evaluate such an expression we simply follow a conversion rule of converting octal to decimal number system
For conversion the following steps are to be proceed
such as 011
here 0 indicate the number is octal
and we are require to convert 11 which is (base 8) to decimal (base 10)
11= 1x8^1+1x8^0
=8+1
=9
文章来源: Why printf output of the variable is different
