CD4 (28)Nov 17, 2009 at 12:14pm UTC
1 #include<stdlib.h>
#include<stdio.h>
#include<string.h>
char *envstrn;
int main( int argc,char *argv[]) {
if (argc==2) {
envstrn=getenv(argv[1]);
printf("%s\n" ,envstrn);
}
return 0;
}
Last edited on Nov 17, 2009 at 12:15pm UTC
jsmith (5804)Nov 17, 2009 at 1:16pm UTC
it returns a null pointer and you aren't checking.
CD4 (28)Nov 17, 2009 at 2:01pm UTC
i also know that its returning a null pointer.. but why... when i normally give
Seph (2)Nov 17, 2009 at 3:12pm UTC
Hello !!if(argc==2) {
envstrn=getenv(argv[1]);
if(envstrn != NULL)
printf("%s\n",envstrn);
else
printf("%s\n", "Environtment variable not found.");
}
CD4 (28)Nov 17, 2009 at 3:58pm UTC
hello seph.. so you mean that otherwise the program is fine.. actually i dint prepare for that NULL because i dint think that OSTYPE wont return a value as it was doing in the console.. Thanks for ur help.. i was great.. :-)
writetonsharma (1181)Nov 18, 2009 at 9:09am UTC
echo itself is a program which takes care of all these things, your program should take care of these errors.
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playing with enviornment variables..