From ea10fc92c547a9194330611f34a680cdef94aa7e Mon Sep 17 00:00:00 2001
From: Matheus Felipe <50463866+matheusfelipeog@users.noreply.github.com>
Date: Sat, 20 Feb 2021 18:51:09 -0300
Subject: [PATCH] Update examples to Python 3 in structure section

---
 docs/writing/structure.rst | 8 ++++----
 1 file changed, 4 insertions(+), 4 deletions(-)

diff --git a/docs/writing/structure.rst b/docs/writing/structure.rst
index 69c9cf616..5d9645acd 100644
--- a/docs/writing/structure.rst
+++ b/docs/writing/structure.rst
@@ -788,7 +788,7 @@ compute x + 1, you have to create another integer and give it a name.
 
     my_list = [1, 2, 3]
     my_list[0] = 4
-    print my_list  # [4, 2, 3] <- The same list has changed
+    print(my_list)  # [4, 2, 3] <- The same list has changed
 
     x = 6
     x = x + 1  # The new x is another object
@@ -822,7 +822,7 @@ most idiomatic way to do this.
     nums = ""
     for n in range(20):
         nums += str(n)   # slow and inefficient
-    print nums
+    print(nums)
 
 **Better**
 
@@ -832,7 +832,7 @@ most idiomatic way to do this.
     nums = []
     for n in range(20):
         nums.append(str(n))
-    print "".join(nums)  # much more efficient
+    print("".join(nums))  # much more efficient
 
 **Best**
 
@@ -840,7 +840,7 @@ most idiomatic way to do this.
 
     # create a concatenated string from 0 to 19 (e.g. "012..1819")
     nums = [str(n) for n in range(20)]
-    print "".join(nums)
+    print("".join(nums))
 
 One final thing to mention about strings is that using ``join()`` is not always
 best. In the instances where you are creating a new string from a pre-determined