windows vista - memory allocation in C -

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i have question regarding memory allocation order. in following code allocate in loop 4 strings. when print addresses don't seem allocated 1 after other... doing wrong or sort of defense mechanism implemented os prevent possible buffer overflows? (i use windows vista).

thank you.

 char **stringarr;  int size=4, i;   stringarr=(char**)malloc(size*sizeof(char*));  (i=0; i<size; i++)     stringarr[i]=(char*)malloc(10*sizeof(char));   strcpy(stringarr[0], "abcdefgh");  strcpy(stringarr[1], "good-luck");  strcpy(stringarr[2], "mully");  strcpy(stringarr[3], "stam");   (i=0; i<size; i++) {   printf("%s\n", stringarr[i]);   printf("%d  %u\n\n", &(stringarr[i]), stringarr[i]);  }

output:

abcdefgh 9650064 9650128

good-luck 9650068 9638624

mully 9650072 9638680

stam 9650076 9638736

typically when request memory through malloc(), c runtime library round size of request minimum allocation size. makes sure that:

  • the runtime library has room bookkeeping information
  • it's more efficient runtime library manage allocated blocks multiples of size (such 16 bytes)

however, these implementation details , can't rely on particular behaviour of malloc().


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