Thread: memory loss again

version gcc 2.95.3
SunOS 5.6

Hi,

why can I not put a printf() at the beginning of a function ?

I do not remember this being a problem, but at the moment I find this does not work

Code:

void function(int parameter)
{
   printf("test");

   int variable;

   variable = 1;

   printf("value in variable is %d\n", variable);

}

whereas this works

Code:

void function(int parameter)
{
   int variable;

   variable = 1;

   printf("test");

   printf("value in variable is %d\n", variable);

}

it appears as though an assigment has to be made before it is allowed, is this true ?


example of error from a make:-

gcc -c telex_filter.c -I../lib -I../common
telex_filter.c: In function `telex_filter_process':
telex_filter.c:86: parse error before `extern'
make: *** [telex_filter.o] Error 1


my reason for this, is to check what is coming into the function, i.e., what values are being passed in,



tia,

well, well

thank you Salem, I never knew that, therefore I can just print out the arguments with a printf() statement placed after the declarations, excellent


ty saravanan.T.S. that won't be necessary, as I'll need to keep these as C programs