c - Reference to a pointer is lost -

this question has answer here:

• assignment of function parameter has no effect outside function 2 answers

i trying understand why statement doesn't work.

char resp[] = "123456789"; void getvalue(char *im) {    im = resp;    printf("\n%s\n",im); }  int main(int argc, char *argv[]) {     char imei[11] = {0};     getvalue(imei);     printf("\nimei: %s\n",imei);     return 0; } 

output:

123456789 imei:  

you can not assign =, use strcpy instead:

#include <stdio.h> #include <string.h>  char resp[] = "123456789"; void getvalue(char *im) {    im = strcpy(im, resp);    printf("\n%s\n",im); }  int main(int argc, char *argv[]) {     char imei[11] = {0};     getvalue(imei);     printf("\nimei: %s\n",imei);     return 0; } 

that's because imei array[11] (not pointer to), if want assign via = can:

#include <stdio.h>  char resp[] = "123456789"; void getvalue(char **im) {    *im = resp;    printf("\n%s\n",*im); }  int main(int argc, char *argv[]) {     char *imei; /* not array pointer */     getvalue(&imei);     printf("\nimei: %s\n",imei);     return 0; }