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Understanding some code

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Started by Stormtrooper December 18, 2008 07:04 PM

13 comments, last by rip-off 15 years, 4 months ago

439

December 19, 2008 11:16 AM

Quote:Original post by Eeyore
Hey all, I just wanted some help understanding some code.
const char *cmd = argv[0] && *argv[0] ? argv[0] : "git-help";

This is logically equivalent to:

const char *cmd;if (argv[0] != NULL){	if (*argv[0] != NULL)	{		cmd = argv[0];	}	else	{		cmd = "git-help";	}}else{	cmd = "git-help";}

That is because of the early-out of the && operator. If the left condition fails, it will never check/execute what's on the right side.

rip-off

11,000

December 19, 2008 11:20 AM

NULL is typically used with pointer values. As such, I would use the NUL character for the second condition:

const char *cmd = "git-help";if (argv[0] != NULL){	if (*argv[0] != '\0')	{		cmd = argv[0];	}}

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phresnel

953

December 20, 2008 02:14 AM

Quote:Original post by rip-off
NULL is typically used with pointer values. As such, I would use the NUL character for the second condition:

Just an addendum: "Should I use NULL or 0?"

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shurcool

439

December 20, 2008 12:59 PM

Quote:Original post by rip-off
NULL is typically used with pointer values. As such, I would use the NUL character for the second condition.

You're absolutely right. However, I was just demonstrating what the logic behind the original and somewhat hard-to-understand statement was, by translating it to an equivalent if/else block. And the original code does in fact check that both argv[0] && *argv[0] have non-0 values, which is equivalent to != NULL if NULL is defined as 0. :)

rip-off

11,000

December 20, 2008 02:22 PM

NULL may be 0, but my point is about intent. When I see something compared with NULL, I'm going to assume that the other thing is a pointer. If I see it compared with '\0' then my assumption is that the other thing is a character. That is all [smile]

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Understanding some code

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