Codes for exercise on data types, and function
parameters. / Answer the Table below by analyzing/ modifying the
code
*/
#include <stdio.h>
#include <string.h>
typedef char String[11];
/*
Lines 29 to 39 are called "function prototypes"
or "function declarations". A function prototype only
contains
the functions's return type, function name and
formal parameters.
Your task is to replace the blank (underline)
with appropriate data type for the formal parameter named
param.
When you are done encoding your answer, go to
the command line, and compile the program
NOTES: the -c is a compiler option which will
compile only (but not link) the C program. There should NOT
be any compilation
error or warning! If
there is an error or warning, then there is something wrong
somewhere. Fix your codes.
If there is no error,
a new file will be produced. This is an object
file.
Do NOT run the
program because it is not a stand-alone exe file!
*/
void Zero( ___ param ); // answer: char * (or char*)
void One( ___ param );
void Two( ___ param );
void Three( ___ param );
void Four( ___ param, char *str );
void Five( ___ param );
void Six( ___ param );
void Seven( ___ param );
void Eight( ___ param );
void Nine( ___ param );
void Ten( ___ param );
int
main()
{
/* Do NOT change any of the
declarations/definitions below. */
char str[11];
// Assume that &str[0] is FE30.
char S[11] = "ABCDEVWXYZ";
// Assume that &S[0]
is FE20.
int A[10] = {1, 3, 5, 2, 4, 9, 8, 6, 7};
// Assume that &A[0] is FDF0.
double M[2][3] = { {1.1, 8.8, -5.5}, {-3.3, 7.7}
}; // Assume that &M[0][0]
is FDC0.
String L[5] = { "Hello!", "How", "are", "you?"};
// Assume that
&L[0] is FD80.
String Z[2][5] = { {"NEKO", "INU", "USAGI",
"TORA", "KUMA"}, // Assume that &Z[0][0] is FD10.
{"CAT", "DOG", "DEER", "TIGER", "BEAR"}
};
strcpy(str, L[0]); // do not change this
strcpy().
/*
The following are
function calls. You need to answer two questions in the
Canvas Quiz for each function call:
Q1: What is the value of the
parameter?
Q2: What is the data type of the
parameter?
For example, for function
call Zero(str+2): FE32 is the value of the parameter, and char * is
its data type.
Explanation: recall from our
recent lecture that acdg to K&R: the name of the array is
snynonymous with the
location (i.e., memory address) of the initial
element (i.e., 1st element).
Based on the given comment in
Line 46: str is synonymous with &str[0] which is assumed above
to be address FE30.
Thus, str + 2 is equal to
FE30 + 2 = FE32.
*/
Zero(str+2);
// Q0 FE32 char
*
One(&A[9]);
Two(S[7]);
Three(*(&A[5]+1));
Four(strcpy(L[4], "Bye!"), "Hello");
Five(*(S+A[5]));
Six(M[1][2]);
Seven(&M[1][2]);
Eight(&L[2][1]);
Nine(Z[1][3]);
Ten(M[1]);
return 0;
}
Answer this table below
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