feature/tyc_in_24_h_part3

helloworld/H11_E1.c

@@ -0,0 +1,44 @@
+//1. Given three integer variables, x = 512, y = 1024, and z = 2048, write a program
+//to print out their left values as well as their right values.
+
+#include <stdio.h>
+
+main() {
+	int x = 512;
+	int y = 1024;
+	int z = 2048;
+
+	printf("x: address=%p, content=%d\n", &x, x);
+	printf("y: address=%p, content=%d\n", &y, y);
+	printf("z: address=%p, content=%d\n", &z, z);
+
+
+//2. Write a program to update the value of the double variable flt_num from 123.45
+//to 543.21 by using a double pointer.
+
+	double flt_num = 123.45;
+	double *ptr_flt_num = &flt_num;
+	*ptr_flt_num = 543.21;
+	printf("flt_num =%f\n", flt_num);
+
+
+//3. Given a character variable ch and ch = 'A', write a program to update the value of
+//ch to decimal 66 by using a pointer.
+
+	char ch = 'A';
+	char *ptr_ch = &ch;
+	*ptr_ch = 66;
+	printf("ch returns %c\n", ch);
+
+//4. Given that x=5 and y=6, write a program to calculate the multiplication of the two
+//integers and print out the result, which is saved in x, all in the way of indirection
+//(that is, using pointers).
+
+	x = 5;
+	y = 6;
+	int *ptrx = &x;
+	int *ptry = &y;
+	*ptrx *= *ptry;
+	printf("x=%d\n", x);
+	return 0;
+}

helloworld/H12_E1.c

@@ -0,0 +1,83 @@
+/*1. Given this character array:
+char array_ch[5] = {'A', 'B', 'C', 'D', 'E'};
+write a program to display each element of the array on the screen.*/
+#include <stdio.h>
+
+main() {
+	char array_ch[5] = {'A', 'B', 'C', 'D', 'E'};
+	int i;
+	for (i=0; i<5; i++){
+		printf("%c", array_ch[i]);
+	}
+	printf("\n");
+
+/*2. Rewrite the program in Exercise 1, but this time use a for loop to initialize the
+character array with 'a', 'b', 'c', 'd', and 'e', and then print out the value of
+each element in the array.*/
+
+	for (i=0; i<5; i++){
+	array_ch[i] = i + 97;
+	printf("%c",array_ch[i]);
+	}
+
+
+/*3. Given this two-dimensional unsized array:
+char list_ch[][2] = {
+'1', 'a',
+'2', 'b',
+'3', 'c',
+'4', 'd',
+'5', 'e',
+'6', 'f'};
+write a program to measure the total bytes taken by the array, and then print out all
+elements of the array.*/
+
+	char list_ch[][2] = {
+		'1', 'a',
+		'2', 'b',
+		'3', 'c',
+		'4', 'd',
+		'5', 'e',
+		'6', 'f'};
+	int total_byte = sizeof(list_ch);
+	int number_of_elements = total_byte / sizeof(char);
+	int imax = number_of_elements / 2;
+	for (i=0; i<imax; i++){
+		printf("\n");
+		for ( int j=0; j<2; j++){
+		printf("%c", list_ch[i][j]);
+		}
+	}
+	printf("\n");
+
+/*4. Rewrite the program in Listing 12.5. This time put a string of characters, I like
+C!, on the screen.*/
+
+	char array_ch_new[10] = {'I',' ','l','i','k','e',' ','C','!','\0'};
+	for (i=0; array_ch_new[i]; i++){
+//	for (i=0; i < 9; i++){
+		printf("%c", array_ch_new[i]);
+		}
+	printf("\n");
+
+/*5. Given the following array:
+double list_data[6] = {
+1.12345,
+2.12345,
+3.12345,
+4.12345,
+5.12345};
+use the two equivalent ways taught in this lesson to measure the total memory
+space taken by the array, and then display the results on the screen.*/
+	double list_data[6] = {
+		1.12345,
+		2.12345,
+		3.12345,
+		4.12345,
+		5.12345};
+	total_byte = sizeof(list_data);
+	printf("The total memory space taken by the array by method I.: %d\n", total_byte);
+	total_byte = sizeof(double) * 6;
+	printf("The total memory space taken by the array by method II.: %d\n", total_byte);
+	return 0;
+}

helloworld/H12_E4_separated.c

@@ -0,0 +1,12 @@
+#include <stdio.h>
+
+main() {
+	int i;
+	char array_ch_new[9] = {'I',' ','l','i','k','e',' ','C','!'};
+//	for (i=0; array_ch_new[i]; i++){
+	for (i=0; i<9; i++){
+		printf("%c", array_ch_new[i]);
+		}
+	printf("\n");
+	return 0;
+}

helloworld/H13_E1.c

@@ -0,0 +1,55 @@
+//1. Given a character array in the following statement,
+//char str1[] = "This is Exercise 1.";
+//write a program to copy the string from str1 to another array, called str2.
+#include <stdio.h>
+#include <string.h>
+main() {
+	setbuf(stdout, NULL);
+	char str1[] = "This is Exercise 1.";
+	char str2[strlen(str1)];
+	strcpy (str2, str1);
+	printf("%s\n", str2);
+
+
+//2. Write a program to measure the length of a string by evaluating the elements in a
+//character array one by one until you reach the null character. To prove you get the
+//right result, you can use the strlen() function to measure the same string again.
+
+	char array_ch_new[10] = {'I',' ','l','i','k','e',' ','C','!','\0'};
+	int i;
+	int length = 0;
+	for (i=0; array_ch_new[i]; i++) {
+		length ++;
+	}
+	printf("The lenght of string is %d\n", length);
+	length = strlen(array_ch_new);
+	printf("The lenght of string is %d\n", length);
+
+//3. Rewrite the program in Listing 13.4. This time, convert all uppercase characters to
+//their lowercase counterparts.
+
+
+	char str[80];
+	int delt = 'a' - 'A';
+	printf("Enter a string less than 80 characters:\n");
+	gets(str);
+	i = 0;
+	while (str[i]) {
+	if ((str[i] >= 'A') && (str[i] <= 'Z'))
+		str[i] += delt; /* convert to lower case */
+		++i;
+	}
+	printf("The entered string is (in lowercase):\n");
+	puts(str);
+
+
+//4. Write a program that uses the scanf() function to read in two integers entered by
+//the user, adds the two integers, and then prints out the sum on the screen.
+
+	int x = 0;
+	int y = 0;
+	printf("Enter two integer:\n");
+	scanf("%d %d", &x, &y);
+	printf("The sum of the numbers is %d", x + y);
+	return 0;
+}

helloworld/H14_E1.c

@@ -0,0 +1,33 @@
+//1. Given the following:
+//• a.) An int variable with block scope and temporary storage
+//• b) A constant character variable with block scope
+//• c) A float local variable with permanent storage
+//• d) A register int variable
+//• e) A char pointer initialized with a null character
+//write declarations for all of them.
+
+//a)
+
+{
+	int x;
+}
+
+//b) 
+
+{
+	const char ch;
+}
+
+//c)
+
+{
+	static float number;
+}
+
+//d)
+
+register int i;
+
+// e)
+
+char *ptr_ch = 0;

helloworld/H14_E2.c

@@ -0,0 +1,36 @@
+//2. Rewrite the program in Listing 14.2. This time, pass the int variable x and the
+//float variable y as arguments to the function_1() function. What do you get on
+//your screen after running the program?
+
+#include <stdio.h>
+
+int x = 1234; /* program scope */
+double y = 1.234567; /* program scope */
+
+void function_1(int x, double y) {
+printf("From function_1:\n x=%d, y=%f\n", x, y);
+}
+
+main() {
+	int x = 4321; /* block scope 1*/
+	function_1(x, y);
+	printf("Within the main block:\n x=%d, y=%f\n", x, y);
+	/* a nested block */
+	{
+		double y = 7.654321; /* block scope 2 */
+		function_1(x, y);
+		printf("Within the nested block:\n x=%d, y=%f\n", x, y);
+	}
+	return 0;
+/* Answer: 
+From function_1:
+ x=4321, y=1.234567
+Within the main block:
+ x=4321, y=1.234567
+From function_1:
+ x=4321, y=7.654321
+Within the nested block:
+ x=4321, y=7.654321
+ */
+
+}

helloworld/H14_E3.c

@@ -0,0 +1,21 @@
+//3. Compile and run the following program. What do you get on the screen, and why?
+#include <stdio.h>
+int main() {
+	int i;
+	for (i=0; i<5; i++){
+		int x = 0;
+		static int y = 0;
+	printf("x=%d, y=%d\n", x++, y++);
+}
+return 0;
+/*
+The screen: 
+x=0, y=0
+x=0, y=1
+x=0, y=2
+x=0, y=3
+x=0, y=4
+
+x is set to 0 every time the program enters the for loop, 
+y has a permanent sorage, so the value saved in y is kept.*/
+}

helloworld/H14_E4.c

@@ -0,0 +1,24 @@
+//4. Rewrite the add_two() function in Listing 14.3 to print out the previous result of
+//the addition, as well as the counter value.
+
+#include <stdio.h>
+/* the add_two function */
+int add_two(int x, int y) {
+	static int counter = 1;
+	static int prev_sum = 0;
+	printf("This is the function call of %d,\n", counter++);
+	printf("The previous result was %d,\n", prev_sum);
+	prev_sum = x + y;
+	return (x + y);
+}
+/* the main function */
+main() {
+	int i, j;
+
+	for (i=0, j=5; i<5; i++, j--)
+	printf("the addition of %d and %d is %d.\n\n", i, j, add_two(i, j));
+	return 0;
+
+//Azért ha őszinte vagyok, ezt a static dolgot nem értem.
+
+}