Pointer as index value in for loop

Question

I'd like to use a few for loops in C and use common index for these. I want to use pointer, because I will be able to free it later.
I thought about something like:

uint8_t *p = malloc(sizeof(uint8_t));

for (*p = 0; *p < 255; *p++)
{
  // Instructions...
}

// A few loops...

free(p);

But this code doesn't work as I want. How can I fix it?

PS I'm a beginner in C's pointers.

In the code as shown, you're not accomplishing anything you couldn't have done by declaring an ordinary variable uint8_t i;. There are uses of dynamic memory allocation, but this isn't it! — Steve Summit
– Steve Summit, Commented Dec 1, 2019 at 13:15
Also, using an 8-bit type as an index variable is usually a bad idea, and is a rather frequent source of subtle bugs. Is there a reason you're trying to use uint8_t here, instead of plain int? — Steve Summit
– Steve Summit, Commented Dec 1, 2019 at 13:17
I used uint8_t because I wanted to save memory. I will check what will happen if I use int, wait a second — dotandl
– dotandl, Commented Dec 1, 2019 at 13:19
In general, for individual variables like that, using a type smaller than int uses more memory, runs more slowly, introduces bugs, and wastes your time. It's a loss in almost every way. My advice to you is: just use int! — Steve Summit
– Steve Summit, Commented Dec 1, 2019 at 13:49

Youssef13 · Accepted Answer · 2019-12-01 13:05:46Z

2

By this line:

uint8_t *p = malloc(sizeof(uint8_t));

You're allocating memory that is suitable to hold ONE uint8_t. You shouldn't touch any other addresses after p because you didn't allocate them. You may need to do the following:

uint8_t *p = malloc(255 * sizeof(uint8_t));

The previous line allocates memory that can hold 255 uint8_t instead of just one. That way, you can access addresses from p to p + 254.

answered Dec 1, 2019 at 13:05

Youssef13

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2 Comments

dotandl Over a year ago

It works, but is too hard to understand for me. I will go back to normal index variable. Thanks!

Youssef13 Over a year ago

malloc simply takes the number of bytes you need from the RAM. You code says to the operating system: "Hey Windows (or linux or mac), give me a memory that can hold uint8_t value". But you actually want 255 values. that's why I multiplied the sizeof(uint8_t) by 255. This way, I'm saying: "Hey Windows, give me memory that can hold 255 uint8_t values.".

dotandl · Accepted Answer · 2019-12-01 13:25:37Z

0

OK, I solved the problem. My code:

uint8_t *p = malloc(sizeof(uint8_t));

for (*p = 0; *p < 255; (*p)++)
{
  // Instructions...
}

// A few loops...

free(p);

Thank you all for your answers!

answered Dec 1, 2019 at 13:25

dotandl

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2 Comments

Steve Summit Over a year ago

I'm glad to know it works, but it would be easier to read, easier to write, faster, use less code, and use less data if you were to simply write uint8_t i; and use i everywhere instead of *p. You are also skating on thin ice here: it is very easy to write a loop that runs forever, when you try to use an 8-bit variable to count to 256. (You're saved by the fact that you're only counting to 255.)

Steve Summit Over a year ago

I said it would "use less code and use less data if you were to simply write uint8_t i;", but you would likely save even more if you changed it to plain int i;.

robsiemb · Accepted Answer · 2019-12-01 15:23:20Z

-1

Syntax of malloc is :

int * p = (*int )malloc(sizeof(int);

You have to first convert the data type into pointer

edited Dec 1, 2019 at 15:23

robsiemb

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answered Dec 1, 2019 at 13:13

Anshal khan

1

1 Comment

Steve Summit Over a year ago

No, casts on malloc's return value are not required, and are disrecommended by most C experts.

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