Declaration not allowed here error in C - c

Question

The following line has the problem int (*f)(int, int) = (argv[2][0] == 'd') , on compiling it says declaration not allowed here . Should the line be declared at the start , any better way of doing this .Any suggestions would be greatly appreciated?
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
int encode(int ch, int key) {
if (islower(ch)) {
ch = (ch-'a' + key) % 26 + 'a';
ch += (ch < 'a') ? 26 : 0;
}
else if (isupper(ch)) {
ch = (ch-'A' + key) % 26 + 'A';
ch += (ch < 'A') ? 26 : 0;
}
return ch;
}
int decode(int ch, int key) {
return encode(ch, -key);
}
int main(int argc, char **argv) {
int ch;
int key;
if (argc < 2) {
printf("USAGE: cipher <integer key> <encode | decode>\n");
printf("Then, just type your text and it will automatically output the en/de crypted text! :)\n");
return 1;
}
key = atoi(argv[1]);
if (key < 1 || key > 25) {
printf("Key is invalid, or out of range. Valid keys are integers 1 through 25.\n");
return 1;
}
int (*f)(int, int) = (argv[2][0] == 'd') ?
decode :
encode;
while (EOF != (ch=getchar()))
putchar(f(ch, key));
return 0;
}

Answer 1

In C (prior to C99), you have to declare variables at the start of a block.
Either compile your code as C99, or change the code so that f is declared at the start of a block.

Answer 2

In c89/90 You must declare all the variables in the starting of the block
But In c99 , You can compile your code with -std=c99 like this:
gcc -Wall -std=c99 test.c -o test.out

Answer 3

Other than the part pointed out by NPE, you can use typedef to create a Function Type. like this:
typedef void FunctionType (int, int); And then use it(as a separate type) to create function pointers.
Makes reading easy.

Answer 4

Should the line be declared at the start
In C89 definitions must occur before any statements in the block. If you do move it, you don't have to move the whole line (and of course you don't want to move the whole line to before the code that checks argv[2] is valid). Just move the definition of f:
int ch;
int key;
int (*f)(int,int);
...
f = (argv[2][0] == 'd') ? decode : encode;
any better way of doing this
It's not necessarily better in this case, but note that the rule is the start of a block, not necessarily the start of a function.
So, you could just write:
{
int (*f)(int, int) = (argv[2][0] == 'd') ?
decode :
encode;
while (EOF != (ch=getchar()))
putchar(f(ch, key));
}
return 0;
You can easily get into arguments about this coding style. Some people think every function should define all its variables up front, and that introducing a block just to define a variable is cluttered and/or confusing. Some people (and especially those who use C++ as well as C) think you should restrict the scope of each variable to as narrow a piece of code as possible, that that defining everything at the start of the function is cluttered and/or confusing. But even they might consider a bare block excessive.