python符號％u和％o

Question

我是新來的編程人員，並試圖弄清楚在不同字母的打印語句中％符號的作用。我理解幾乎所有這些接受％u所做的事情。它似乎只是打印538作爲一個整數。我看到前面的打印語句用'u'打印在unicode中，但我不知道它是否適用於％u。

print "In honor of the election I present %d" % 538.0 # integer 
print "In honor of the election I present %o" % 538.0 # octal 
print "In honor of the election I present %u" % 538.0 # ? 
print "In honor of the election I present %x" % 538.0 # lowercase hexadecimal 
print "In honor of the election I present %X" % 538.0 # uppercase hexadecimal 
print "In honor of the election I present %e" % 538.0 # exponential 
print "In honor of the election I present %i" % 538.0 # integer 

輸出如下：

In honor of the election I present 538 
In honor of the election I present 1032 *emphasized text* 
In honor of the election I present 538 *emphasized text* 
In honor of the election I present 21a 
In honor of the election I present 21A 
In honor of the election I present 5.380000e+02 
In honor of the election I present 538 

我也有與%o有點麻煩，這個號碼。我剛剛瞭解到八進制打印是什麼，我想它會輸出132（538 --> 8^3 = 512 *(1) + 26, 8^1 = 8*(3) + 2, 8^0 = 1*(2)），但輸出是1032。 0從哪裏來？

Answer 1

the docs從，%u是

作廢類型 - 它是相同的'd'。

%o告訴print解釋538作爲基10，並將其轉換爲八進制數。 538基座10（538 ）是1032以八進制（1032 ）：

1 * 8^3 + 0 * 8^2 + 3 * 8^1 + 2 * 8^0 
= 512 + 0 + 24 + 2 
= 538 

它顯示1032，因爲這些都是8 ^Ñ適當的係數。 0對應於8 。如果你離開它，你就會有132 = 1 * 64 + 3 * 8 + 2 = 90 ，而不是538

所以，沒有什麼奇怪的存在。