Pointer confusion

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# 1  
Old 02-16-2014
Pointer confusion

Here are two programs that pass a pointer to a variable but behave differently. Shouldnt the i in second program be 0 after the function call?


void changeI(int *i)
 *i = 10;

int main(void)
 int i=5;
 printf("%d before\n", i);
 printf("%d after\n", i);
 return 0;

chlsvnc01> ./temp
5 before
10 after

chlsvnc01> cat prttst3.c

void copy_array(int * destArray, int * srcArray, int *i)
  *destArray++ = *srcArray++;
 printf("%d is after copy\n", *i);
 return ;

int main (void)
 int i=30;
 int s[30] = {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29};
 int d[30];

 printf("i before = %d\n",i);
 printf("i after = %d\n",i);

 return 0;

chlsvnc01> ./prttst3
i before = 30
0 is after copy
i after = 30
0       1       2       3       4       5       6       7       8       9       10      11      12      13      14      15      16      17      18      19      20      21      22      23      24      25      26      27      28      29

Last edited by bartus11; 02-16-2014 at 12:02 PM.. Reason: Please use [code][/code] tags.
# 2  
Old 02-16-2014
The problem is that:


decrements the pointer, not the pointed-to value. I.e. it's equivalent to:


What you really want is:


(On my machine at least, your program worked by accident simply because of the location of i and s on the stack.)
This User Gave Thanks to JohnGraham For This Post:
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