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An example project, for my column named "Core Golang - 36 lessons"

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golang_puzzlers's Issues

sign长度应该等于max1

Golang_Puzzlers/src/puzzlers/article22/q0/demo58.go

Line 28 in aa99ef9

sign := make(chan struct{}, 2)

下边的61行也应该循环max1次,从sign中取值

一个小小的失误

# cat demo21.go 
package main

func main() {
	// 示例1。
	ch1 := make(chan int, 1)
	ch1 <- 1
	//ch1 <- 2 // 通道已满，因此这里会造成阻塞。

	// 示例2。
	ch2 := make(chan int, 1)
	//elem, ok := <-ch2 // 通道已空，因此这里会造成阻塞。
	//fmt.Printf("%v, %v", elem, ok)
	//_, _ = elem, ok
	ch2 <- 1

	// 示例3。
	var ch3 chan int
	//ch3 <- 1 // 通道的值为nil，因此这里会造成永久的阻塞！
	//<-ch2 // 通道的值为nil，因此这里会造成永久的阻塞！   ---->  这个ch2是不是要改成ch3呢
	_ = ch3
}

Some questions about panic

Summary

在 demo48 中，说明了控制权会一级一级地沿着调用栈的反方向传播至顶端。我就把代码改了改，变成了python，发现是和go的传播方式是反着的。这两种方式只是设计思路的差异吗？还是说，因为我这段 python 是面向过程的，所以，就展示的是这样？

求郝大或者其他大神解答

代码展示

#!/usr/bin/python
def main():
	print "Enter function main."
	caller1()
	print "Exit function main."

def caller1():
	print "Enter function caller1."
	caller2()
	print "Exit function caller1."

def caller2():
	print "Enter function caller2."
	s1 = [0, 1, 2, 3, 4]
	e5 = s1[5]
	print "Exit function caller2. " 

main()

这些是打印结果：
Enter function main.
Enter function caller1.
Enter function caller2.
Traceback (most recent call last):
File "demo48.py", line 18, in
main()
File "demo48.py", line 4, in main
caller1()
File "demo48.py", line 9, in caller1
caller2()
File "demo48.py", line 15, in caller2
e5 = s1[5]
IndexError: list index out of range