Why "a+++++b" can not be compiled in gcc, but "a+++b", "a++ + ++b", and "a+++ ++b" can be? [duplicate]

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Please help me understanding the error a+++++b in C

Here is is the sample code, why "a+++++b" can not be compiled , but others can be?

#inclu开发者_JS百科de <stdio.h>
#include <stdlib.h>

int main(int argc, char **argv)
{
    int a = 0;
    int b = 0;
    int c = 0;
    c = a+++b;
    printf("a+++b is: %d\n", c);

    c = a = b = 0;
    c = a++ + ++b;
    printf("a++ + ++b is: %d\n", c);

    c = b = a = 0;
    c = a+++ ++b;
    printf("a+++ ++b is: %d\n", c);

    c = b = a = 0;
    c = a+++++b;      // NOTE: Can not be compiled here.
    printf("a+++++b is: %d\n", c);

    return 0;
}

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That's because a+++++b is parsed as a ++ ++ + b and not as a ++ + ++ b[C's tokenizer is greedy]. a++ returns an rvalue and you cannot apply ++ on an rvalue so you get that error.

a+++b; // parsed as a ++ + b
a+++ ++b; // parsed as a ++ + ++ b

Read about Maximal Munch Rule.

---

The compiler is greedy so your expression

a+++++b

will be understood as

a++ ++ +b

---

The + operators cascade ... with a+++++b, there is no l-value (memory addressable value) to add against after the addition operations are cascaded.

Put another way, a+++b is the same as (a++) + b. That's a valid operation. The same is true with a+++ ++b which equates to (a++) + (++b). But with a+++++b, you don't get that via the C-parser. To the parser it looks like ((a++)++) + b, and since (a++) returns a temp, that's not an l-value that can be incremented again via the ++ operator.

---

# include <stdio.h>
# include <stdlib.h>

int main(int argc, char **argv)
{
 int a = 0;
 int b = 0;
 int c = 0;
 c = a+++b;
 printf("a+++b is: %d\n", c);

 c = a = b = 0;
 c = (a++)+(++b);
 printf("a++ + ++b is: %d\n", c);

 c = b = a = 0;
 c = (a++)+(++b);
 printf("a+++ ++b is: %d\n", c);

 c = b = a = 0;
 c = (a++)+(++b);     
 printf("a+++++b is: %d\n", c);

 return 0;
}