malloc and how properly to use-it

hi friends,

I am reading this book: Amazon.com: C Programming for the Absolute Beginner (For the Absolute Beginner (Series).) (9781931841528): Michael Vine: Books

And to be on right track I test things, think a bit... and when I have some questions I try to post here sine we have many advance users...

The author claims this things about malloc, which I would like to hear your opinion.

1. Basic malloc() use is demonstrated in the below.

Code:

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char *name;
name = malloc(80);

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2. The preceding program’s use of malloc() is not quite complete because some C compilers may require that you perform type casting when assigning dynamic memory to a variable.

Code:

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char *name;
name = (char *) malloc(80);

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3. Better yet, we should be more specific when creating dynamic memory by explicitly telling the system the size of the data type for which we are requesting memory.

Code:

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char *name;
name = (char *) malloc(80 * sizeof(char));

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Is the third way the right way to use malloc always? Do you agree... Are you personally using it like that? Please let me know further, I really would like to hear your valuable opinion!

2. No C compilers require that you perform typecasting in this situation -- and in fact doing so is more likely wrong than right in C (since it is used to cover a lack of #include files). However, all C++ compilers do require it.

3. Sort of. sizeof(char) is guaranteed to be 1 and therefore this is rather useless. But in general, you will often see things like:

Code:

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name = malloc(80 * sizeof(*name));

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where instead of using the typename you specify "80 of the things name points to" -- so if this changes, this line will track that change.

Thanks, tabstop!

Quote:

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Originally Posted by tabstop

2. No C compilers require that you perform typecasting in this situation -- and in fact doing so is more likely wrong than right in C (since it is used to cover a lack of #include files). However, all C++ compilers do require it.

3. Sort of. sizeof(char) is guaranteed to be 1 and therefore this is rather useless. But in general, you will often see things like:

Code:

---

name = malloc(80 * sizeof(*name));

---

where instead of using the typename you specify "80 of the things name points to" -- so if this changes, this line will track that change.

---

Dang, ya learn something new every day...
Thanks tabstop... noted for future!

Quote:

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Originally Posted by tabstop

2. No C compilers require that you perform typecasting in this situation -- and in fact doing so is more likely wrong than right in C (since it is used to cover a lack of #include files). However, all C++ compilers do require it.

3. Sort of. sizeof(char) is guaranteed to be 1 and therefore this is rather useless. But in general, you will often see things like:

Code:

---

name = malloc(80 * sizeof(*name));

---

where instead of using the typename you specify "80 of the things name points to" -- so if this changes, this line will track that change.

---

3. Why we're not performing typecasting in this situation, as?

Code:

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name = (int *) malloc(80 * sizeof(*name));

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You don't need to.

Question 7.7

Quote:

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Originally Posted by whiteflags

You don't need to.

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Even in C++?

Quote:

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Originally Posted by c_lady

Even in C++?

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This is the C programming forum, and whiteflags' answer pertains to C. tabstop's answer actually correctly covers both C and C++, but for some reason you trust whiteflags more than tabstop ;)

Quote:

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Originally Posted by laserlight

This is the C programming forum, and whiteflags' answer pertains to C. tabstop's answer actually correctly covers both C and C++, but for some reason you trust whiteflags more than tabstop ;)

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Well, who wouldn't?

I trust everyone in this forum ;) starting first from tabstop and continuing by you laserlight...