I see code like ...

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I see code like ...

I see code like
char *p = malloc(strlen(s) + 1);
strcpy(p, s);

Shouldn't that be malloc((strlen(s) + 1) * sizeof(char))?

✍: Guest

It's never necessary to multiply by sizeof(char), since sizeof(char) is, by definition, exactly 1. (On the other hand, multiplying by sizeof(char) doesn't hurt, and in some circumstances may help by introducing a size_t into the expression;

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💬 2020-05-27 james: very helpful thanks

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I see code like ...
I see code like char *p = malloc(strlen(s) + 1); strcpy(p, s); Shouldn't that be malloc((strlen(s) + 1) * sizeof(char))? It's never necessary to multiply by sizeof(char), since sizeof(char) is, by definition, exactly 1. (On the other hand, multiplying by sizeof(char) doesn't hurt, and in some circum... 2016-06-21, 2799👍, 0💬

I im trying to declare a pointer and allocate some space for it
I'm trying to declare a pointer and allocate some space for it, but it's not working. What's wrong with this code? char *p; *p = malloc(10); The pointer you declared is p, not *p. When you're manipulating the pointer itself (for example when you're setting it to make it point somewhere), you just us... 2016-06-10, 2747👍, 0💬

How can I dynamically allocate arrays?
How can I dynamically allocate arrays? The equivalence between arrays and pointers allows a pointer to malloc'ed memory to simulate an array quite effectively. After executing #include <stdlib.h> int *dynarray; dynarray = malloc(10 * sizeof(int)); (and if the call to malloc succeeds), you can... 2016-04-23, 2672👍, 0💬

I wrote a little wrapper around malloc ...
I wrote a little wrapper around malloc, but it doesn't work: #include lt;stdio.h> #include lt;stdlib.h> mymalloc(void *retp, size_t size) { retp = malloc(size); if(retp == NULL) { fprintf(stderr, "out of memory\n"); exit(EXIT_FAILURE); } } Are you sure the function initialized what you thought it di... 2016-06-10, 2616👍, 0💬

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