C maximal munch strategy (最大一口策略) - c = a+++b; <==> c = a++ + b;

1. c = a+++b; <==> c = a++ + b;

c = a+++b;
等同于
c = a++ + b;

对于 c = a+++b; 语句，程序员的意图可能是 c = a + ++b;，也可能是 c = a++ + b;。
ANSI C 规定了一种逐渐为人所熟知的 maximal munch strategy (最大一口策略)。这种策略表示如果下一个标记有超过一种的解释方案，编译器将选取能组成最长字符序列的方案。c = a+++b; 将被解析为 c = a++ + b;。

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1.1 Example

//============================================================================
// Name        : main
// Author      : Yongqiang Cheng
// Version     : Version 1.0.0
// Copyright   : Copyright (c) 2019 Yongqiang Cheng
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <stdio.h>

int main(int argc, char *argv[])
{
	int a = 0, b = 0, c = 0, d = 0;

	printf("argc = %d\n", argc);

	for (int idx = 0; idx < argc; ++idx)
	{
		printf("argv[%d] --> %s\n", idx, argv[idx]);
	}

	printf("argv[argc] = %p\n\n", (void*)argv[argc]);

	a = 3;
	b = 6;
	c = 0;

	c = a+++b;
	printf("c = a+++b; => c = %d\n", c);
	printf("a = %d\n", a);
	printf("b = %d\n\n", b);

	a = 3;
	b = 6;
	d = 0;

	d = a++ + b;
	printf("d = a++ + b; => d = %d\n", d);
	printf("a = %d\n", a);
	printf("b = %d\n\n", b);

	return 0;
}

argc = 1
argv[0] --> D:\visual_studio_workspace\yongqiang\Debug\yongqiang.exe
argv[argc] = 00000000

c = a+++b; => c = 9
a = 4
b = 6

d = a++ + b; => d = 9
a = 4
b = 6

请按任意键继续. . .

1.2 Example

//============================================================================
// Name        : main
// Author      : Yongqiang Cheng
// Version     : Version 1.0.0
// Copyright   : Copyright (c) 2019 Yongqiang Cheng
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <stdio.h>

int main(int argc, char *argv[])
{
	int a = 0, b = 0, c = 0, d = 0;

	printf("argc = %d\n", argc);

	for (int idx = 0; idx < argc; ++idx)
	{
		printf("argv[%d] --> %s\n", idx, argv[idx]);
	}

	printf("argv[argc] = %p\n\n", (void*)argv[argc]);

	a = 3;
	b = 6;
	c = 0;

	printf("a = %d\n", a);
	printf("b = %d\n", b);
	printf("c = %d\n", c);

	c = a+++b;

	printf("c = a+++b; => c = %d\n", c);
	printf("a = %d\n", a);
	printf("b = %d\n\n", b);

	a = 3;
	b = 6;
	d = 0;

	printf("a = %d\n", a);
	printf("b = %d\n", b);
	printf("d = %d\n", d);

	d = a++ + b;

	printf("d = a++ + b; => d = %d\n", d);
	printf("a = %d\n", a);
	printf("b = %d\n\n", b);

	return 0;
}

argc = 1
argv[0] --> D:\visual_studio_workspace\yongqiang\Debug\yongqiang.exe
argv[argc] = 00000000

a = 3
b = 6
c = 0
c = a+++b; => c = 9
a = 4
b = 6

a = 3
b = 6
d = 0
d = a++ + b; => d = 9
a = 4
b = 6

请按任意键继续. . .

2. c = a+++++b; <!=> c = a++ + ++b;

c = a+++++b; 按照前面的策略将被解析为 c = a++ ++ + b;，这将引起一个编译错误，错误信息是 ++ 操作符迷失于空格间。即使编译器能够推断 (从理论上说) 惟一有效的编排方式是 c = a++ + ++b;，它还是会出现编译错误。

2.1 error C2105: '++' needs l-value

//============================================================================
// Name        : main
// Author      : Yongqiang Cheng
// Version     : Version 1.0.0
// Copyright   : Copyright (c) 2019 Yongqiang Cheng
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <stdio.h>

int main(int argc, char *argv[])
{
	int a = 0, b = 0, c = 0;

	printf("argc = %d\n", argc);

	for (int idx = 0; idx < argc; ++idx)
	{
		printf("argv[%d] --> %s\n", idx, argv[idx]);
	}

	printf("argv[argc] = %p\n\n", (void*)argv[argc]);

	a = 3;
	b = 6;
	c = 0;

	printf("a = %d\n", a);
	printf("b = %d\n", b);
	printf("c = %d\n", c);

	c = a+++++b;

	printf("c = a+++++b; => c = %d\n", c);
	printf("a = %d\n", a);
	printf("b = %d\n\n", b);

	return 0;
}

1>------ Rebuild All started: Project: yongqiang, Configuration: Debug Win32 ------
1>  foreverstrong.c
1>d:\visual_studio_workspace\yongqiang\yongqiang\foreverstrong.c(32): error C2105: '++' needs l-value
========== Rebuild All: 0 succeeded, 1 failed, 0 skipped ==========

2.2 c = a++ + ++b;

//============================================================================
// Name        : main
// Author      : Yongqiang Cheng
// Version     : Version 1.0.0
// Copyright   : Copyright (c) 2019 Yongqiang Cheng
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <stdio.h>

int main(int argc, char *argv[])
{
	int a = 0, b = 0, c = 0;

	printf("argc = %d\n", argc);

	for (int idx = 0; idx < argc; ++idx)
	{
		printf("argv[%d] --> %s\n", idx, argv[idx]);
	}

	printf("argv[argc] = %p\n\n", (void*)argv[argc]);

	a = 3;
	b = 6;
	c = 0;

	printf("a = %d\n", a);
	printf("b = %d\n", b);
	printf("c = %d\n", c);

	c = a++ + ++b;

	printf("c = a+++++b; => c = %d\n", c);
	printf("a = %d\n", a);
	printf("b = %d\n\n", b);

	return 0;
}

argc = 1
argv[0] --> D:\visual_studio_workspace\yongqiang\Debug\yongqiang.exe
argv[argc] = 00000000

a = 3
b = 6
c = 0
c = a+++++b; => c = 10
a = 4
b = 7

请按任意键继续. . .