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Can someone explain to me why my call to malloc with a string size of 6 returns a sizeof of 4 bytes? In fact, any integer argument I give malloc I get a sizeof of 4. Next, I am trying to copy two strings. Why is my ouput of the copied string (NULL)? Following...
Started by John on
, 8 posts
by 8 people.
Answer Snippets (Read the full thread at stackoverflow):
Char * str = "string"; /* char * copy = malloc(sizeof(str) + 1); Oops */ char * copy = malloc(strlen(str) + 1); charYou're getting....
To access the beginning of the string:
Edit to repair malloc/sizeof issue - thanks CL.
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Someone here recently pointed out to me in a piece of code of mine I am using
char* name = malloc(256*sizeof(char)); // more code free(name);
I was under the impression that this way of setting up an array was identical to using
char name[256];
and that...
Started by faceless1_14 on
, 7 posts
by 7 people.
Answer Snippets (Read the full thread at stackoverflow):
Int *p = (int *) malloc....
; }
Break down your statement
char* name = malloc(256*sizeof(char)); // one statement char *name; // Step 1 declare pointer to character name = malloc(256*sizeof(char)); // assign address to pointer.
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Hello,
Please consider:
int i; char other[4096][16]; i = sizeof(other[0]);
i = 16... which is okay!
Now please consider:
typedef struct T_Lb Lb; struct T_Lb { char** m_Buffer; }; char *LstB1[2] = {"Default", "Stop"}; void LB_NewLb(Lb* pLB, char** mm)...
Started by Viorel_ MVP on
, 7 posts
by 1 people.
Answer Snippets (Read the full thread at microsoft):
Is there any way to get the "16" by ... .
Is there any way to get the "16" by using sizeof is ....
So my question is ....
I] = (char*) malloc(sizeof(char) * 16);
It's sizeof(char) * 16 - you just passed it to mallocwhich is 16.
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Posted 08 May 2012 - 07:48 AM
Hey!
I am wondering what happens if I lets say use 64bit integers on my computer?
As we all know sizeof(int) is not always 4 - it is computer dependent. However CUDA GPUs use 32 bit int - sizeof(int) = 4.
What I am wondering...
Started by Lumpizaver on
, 2 posts
by 2 people.
Answer Snippets (Read the full thread at nvidia):
Posted 15 May 2012 - 08:21 PM
In fact sizeof(int) is *NOT* computer-dependent, but purely compiler.
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I have the following instance method (adapted from Listing 3-6 of the Event Handling section in the iPhone Application Programming Guide):
- (CGPoint)originOfTouch:(UITouch *)touch { CGPoint *touchOriginPoint = (CGPoint *)CFDictionaryGetValue(touchOriginPoints...
Started by Shaun Inman on
, 5 posts
by 5 people.
Answer Snippets (Read the full thread at stackoverflow):
Approach, you have to find a place to free those malloc -ed memory when the values are not needed.
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Could you explain following code?
str = (char *) malloc (sizeof(char) * (num+1));
What's is malloc? Why num + 1?
Started by ffffff on
, 12 posts
by 12 people.
Answer Snippets (Read the full thread at stackoverflow):
For example:
int * p = (int *) malloc (25 * sizeof(int)) ;
Then p points to a memory where youmalloc is for memory....
The first problem is parsing the code
Welcome into the Twilight Zone str = (char *) malloc (sizeof as an array.
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Possible Duplicate:
newbie questions about malloc and sizeof
I am trying to read strings into a program. When I noticed that the strings were sometimes being corrupted, I tried the following code:
void *mallocated = malloc(100); printf("sizeof(mallocated...
Started by Michael Dickens on
, 6 posts
by 6 people.
Answer Snippets (Read the full thread at stackoverflow):
What you probably want is
char (*s)[100] = malloc( sizeof( char[100] ) ); printfBecause the size of the "....
Still!
To know, possibly on the heap.
Or 8 on 64-bit architectures) char *p = malloc(100); // sizeof p == 4 (or 8).
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Recently saw someone commending another user on their use of sizeof var instead of sizeof(type). I always thought that was just a style choice. Is there any significant difference? As an example, the lines with f and ff were considered better than the...
Started by jmucchiello on
, 7 posts
by 7 people.
Answer Snippets (Read the full thread at stackoverflow):
These:
foo **ff = malloc(count * sizeof *ff); foo **gg = malloc(sizeof(foo*) * count);
..but whatIf the type of the variable is changed, the sizeof will not require changing if the variable is the argument, rather than....
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We recently received a report that our application will occasionally fail to run. I tracked down the problem code to this:
struct ARRAY2D { long[] col; } int numRows = 8 ; int numCols = 300; array = (ARRAY2D*) malloc(numRows * numCols * sizeof(long))
...
Started by Brian on
, 5 posts
by 5 people.
Answer Snippets (Read the full thread at stackoverflow):
ARRAY2D() { memset(col, 0, sizeof(col)); } }; // no need to malloc std::unordered_map<(numCols * sizeof(long*)); for (int i = 0; i < numCols; i++) array[i] = (long*) malloc(numRows * sizeof *rows....
Not necessary.
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If I use malloc in my code:
int *x = malloc(sizeof(int));
I get this warning from gcc :
new.c:7: warning: implicit declaration of function ‘malloc’ new.c:7: warning: incompatible implicit declaration of built-in function ‘malloc’
I'm new to C. Am I doing...
Started by Lucas McCoy on
, 6 posts
by 6 people.
Answer Snippets (Read the full thread at stackoverflow):
If you don't do that, the compiler thinks you want to define your own function named malloc ....
To add typecast in addition to include
#include <stdlib.h> int *x = (int*)malloc(sizeof(int includes malloc 's declaration.
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