C Language, question, malloc in a function

question, malloc in a function

Author	Message
tqu.. #1 / 6	question, malloc in a function Hey all- If I allocate space to a local variable with malloc() in a function, and return() from the function without free() -ing the malloc() -ed memory, does the memory return to the heap as the variable passes out of scope? For example: #include<stdio.h> #include<string.h> char substr(char , int, int); main(int argc, char argv[]) { char thing[]="abcdefg"; printf("\n%s",substr(thing,1,3)); Quote: } char substr(char source, int start, int length) { char dest; dest=malloc((length+1)*sizeof(char)); strcpy(dest, ""); strncpy(dest,&source[start-1],length); dest[length]='\0'; return(dest); Quote: } The above code returns a substring of a given string. It works, I am just wondering whether overuse will do bad things. Thanks for any insights, respectfully,
Mon, 02 Nov 1998 03:00:00 GMT

John Winte #2 / 6	question, malloc in a function Quote: >Hey all- >If I allocate space to a local variable with malloc() in a function, >and return() from the function without free() -ing the malloc() -ed memory, >does the memory return to the heap as the variable passes out of scope? No. Quote: >For example: >#include<stdio.h> >#include<string.h> >char substr(char , int, int); >main(int argc, char argv[]) >{ >char thing[]="abcdefg"; >printf("\n%s",substr(thing,1,3)); >} >char substr(char source, int start, int length) >{ >char dest; >dest=malloc((length+1)*sizeof(char)); >strcpy(dest, ""); >strncpy(dest,&source[start-1],length); >dest[length]='\0'; >return(dest); >} >The above code returns a substring of a given string. It works, I am just >wondering whether overuse will do bad things. Yes, using it will do bad things (specifically, a memory leak), but even if the behaviour were as suggested in your first paragraph, it would still do bad things because you'd be returning a pointer to some memory which had been returned to the heap. You could make it the responsibility of the caller to free off the memory, or use a static buffer in the sub-routine, or an array of static buffers used cyclically. John -- John Winters. Wallingford, Oxon, England.
Wed, 04 Nov 1998 03:00:00 GMT

Mike Rubenste #3 / 6	question, malloc in a function Quote: > Hey all- > If I allocate space to a local variable with malloc() in a function, > and return() from the function without free() -ing the malloc() -ed memory, > does the memory return to the heap as the variable passes out of scope? > For example: > #include<stdio.h> > #include<string.h> > char substr(char , int, int); > main(int argc, char argv[]) > { > char thing[]="abcdefg"; > printf("\n%s",substr(thing,1,3)); > } > char substr(char source, int start, int length) > { > char dest; > dest=malloc((length+1)*sizeof(char)); > strcpy(dest, ""); > strncpy(dest,&source[start-1],length); > dest[length]='\0'; > return(dest); > } > The above code returns a substring of a given string. It works, I am just > wondering whether overuse will do bad things. It will eat memory if the user does not free() the string returned when done with it. Memory allocated with mallc() is freed only when you explicitly do so or at program termination (the standard does not actually guarantee the latter, but any reasonable operating system will do it for you). Michael M Rubenstein
Wed, 04 Nov 1998 03:00:00 GMT

Lawrence Kirb #4 / 6	question, malloc in a function Quote: >Hey all- >If I allocate space to a local variable with malloc() in a function, Be clear that you are talking about 2 objects/varioables here. Firstly there is the pointer which may or may not be a local variable, and secondly there is the object you created with malloc which is not a local variable. Quote: >and return() from the function without free() -ing the malloc() -ed memory, >does the memory return to the heap as the variable passes out of scope? If you have an automatic (local) pointer that is destroyed when it passes out of scope. The malloc'd variable itself has no scope since it has no identifier/name. It has a 'duration' i.e. it lasts until it is freed. If you destroy the pointer without first freeing the malloc'd variable it remians allocated but you have no way of getting at it. This is called a memory leak. Quote: >For example: >#include<stdio.h> >#include<string.h> >char substr(char , int, int); >main(int argc, char argv[]) >{ >char thing[]="abcdefg"; >printf("\n%s",substr(thing,1,3)); >} >char substr(char source, int start, int length) >{ >char dest; >dest=malloc((length+1)*sizeof(char)); >strcpy(dest, ""); This strcpy is redundant. If it was needed it would be better to write it as: dest[0] = '\0'; Quote: >strncpy(dest,&source[start-1],length); >dest[length]='\0'; Consider replacing these three lines with: dest[0] = '\0'; strncat(dest,&source[start-1],length); Quote: >return(dest); >} >The above code returns a substring of a given string. It works, I am just >wondering whether overuse will do bad things. If you call substr repeatedly without freeing the variable that malloc creates then you will slowly eat through your available memory. You could in this case free it in the caller after you had used it. However with this sort of code it is normally better to pass an extra argument of a pointer to a buffer created in the caller. -- ----------------------------------------- -----------------------------------------
Wed, 04 Nov 1998 03:00:00 GMT

@#$%! #5 / 6	question, malloc in a function : If I allocate space to a local variable with malloc() in a function, : and return() from the function without free() -ing the malloc() -ed memory, : does the memory return to the heap as the variable passes out of scope? (Assuming the local variable was the only place the mallocked pointer was stored, when the local goes out of the scope without freeing the pointer, the block becomes 'garbage.') In ordinary cases: NO. C does not require the compiler or runtime to support or provide any kind of garbage collection. And many features of the language make complete and safe garbage collection impossible. There are garbage collecting versions of C around, but you would probably know it if you had one. -- the Queen who straits, the Queen of strife;\| Cupertino, California with gasp of death or gift of breath \| (xxx)xxx-xxxx 95015 she brings the choice of birth or knife. \| I don't use no smileys
Wed, 04 Nov 1998 03:00:00 GMT

Bernardo Rein #6 / 6	question, malloc in a function Quote: > Hey all- > If I allocate space to a local variable with malloc() in a function, > and return() from the function without free() -ing the malloc() -ed memory, > does the memory return to the heap as the variable passes out of scope? Sorry, I have no time to check the code: The answer is NO. The pointer is lost, since it is a local variable, but the heap space allocated, remains allocated: That's memory leakage.
Mon, 09 Nov 1998 03:00:00 GMT

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