08-00-Strings of characters - Intro
Defining a string
How to define a variable to hold names of people with a maximum size name of 10
A string of characters is really an array of single characters. That means you need to define an array to hold 10 individual characters the same way as you would define an array that holds 10 individual integers or floats.
charname [10]-- accept this for now …we will fix it later on page 3
This means 10 character size locations (4 bytes) are set up to store this information. The 10 elements of the character array as follows:
If we store a name like MOLIBDENUM it would appear as follows
M / O / L / I / B / D / E / N / U / MIf the next time we store only 4 letters, the character array would “theoretically” look like this
T / A / R / R / B / D / E / N / U / MWhen we printed out the name, we would expect to print TARR but it came out as TARRBDENUM
The simple reason for that is there is no character that tells the printf where to stop printing.
08-01.c - Try this programcut and paste it and run it.
This program demonstrates a 10 character string
#include <stdio.h>
/* 08-01.c - demonstrate getting and printing a string of characters */
main () {
charname [10];
printf("\n\n\n");
printf("Enter a 10 character name : ");
scanf ("%s", name);
printf("\nYou entered [%s]\n", name);
} // END OF MAIN
Do the following:
1Run the same program, but enteronly 5 characters.
2Run the program but enter 12 characters
What happened?
The 10 worked ok, the 5 worked AND so did the 12 character entry. We defined the length of the array as 10, but it still handled 12.
*** When you entered 5, 10 or 12 characters you also pressed the enter key. That means you actually entered 6, 11 or 13 characters.
0 / 1 / 2 / 3 / 4 / 5 / 6 / 7 / 8 / 9 / 10 / 11 / 12 / 13m / o / l / i / b / d / e / n / u / m / \0 / 1st entry into memory
1 / 2 / 3 / 4 / 5 / \0 / e / n / u / m / \0 / 2nd entry in same memory location
L / o / n / g / n / a / m / e / O / f / m / o / r / \0 / 3rd sample of long data
How does printf know when to stop on printout?
The printf function looks for an ending character to tell it when to stop. That character is known as a NULL character to end the string of characters.
NULL means the absence of any value. It does NOT mean zero. Zero has a value of zero. You can’t have a bit pattern that has nothing in it so the bit pattern is as follows:
The bit pattern for 0 is0010 1000 or DEC 48 or HEX 30
The bit pattern for 1 is0010 1001 or DEC 49 or HEX 31
The bit pattern for NUL is0000 0000 or DEC 0 or HEX 00 <== note the character is called NUL
but in English it is written as NULL
This means IF the maximum input can be 10 meaningful characters then you must leave an extra space for the NULL or ‘\0’ character.
Therefore, we should define the variable as follows:
charname [11]remember on page 1, I said I would fix it
The array above would then look as follows
M / O / L / I / B / D / E / N / U / M / \0Whenyou next insert a new value the \0 is added at the end
T / A / R / R / \0 / D / E / N / U / M / \0Notice that the previous characters are still there but the second printf will stop at the first \0
Demonstrate the NULL effect
OR
How to demonstrate what it holds in the array
#include <stdio.h>
/* Program does
1 Demonstrate getting and printing a string of characters and the effect of the NULL character
2 Print out what is in the locations using a for loop*/
voidpause (void);
int main () {
charname [11];// allow 10 plus null
int x;
printf("\n\n\n");
printf("Enter a 10 character name : ");
scanf ("%s", name);
printf("\nYou entered these 10 characters [%s]\n", name);
pause();
printf("\nEnter a 5 character name : ");
scanf ("%s", name);
printf("\nYou entered these 5 characters [%s]\n", name);
for (x=0; x<11;x++) // this will print out every position
printf (“\nIn position %d is … %c ”, x, name[x]);
printf(“\n\nLook at the NON-printable character in position 5\n\n”);
return 0;
}
voidpause (void){ // this function allows me to pause on the screen
while (getchar() != '\n'); // clears the buffer looks for ENTER key
printf ("\n\n\nPause .... hit any key\n\n\n");
while (getchar() != '\n'); // clears the buffer
}
How to load data into an array of characters
Hard coded
int main () {
charname [11]= “Molibdenum”; this is one way of hard coding it
OR .. just like you load integer arrays you do it 1 value at a time with SINGLE quotes
int main () {
charname [11]= {‘M’,’o’,’l’, … ‘m’}; another
with scanf()
Notice the scanf
scanf ("%s", name);
%s – a new placeholder to define a string input of any length
name– Notice that there is not an & in front of the variable name. This is because scanf is looking for an address and the name of an array isthe beginning address of an array
I also found that the &name will work just as well since it is also the start of the array
To get the start of one of the parts of an array you need to do &name[3]
How to limit input to the variable
scanf (“%5s”, name)
This will only insert 5 of the letters into the variable name. If you typed in 15 characters, only the first 5 will go into the array leaving the remainder in the buffer.
Note, this may cause problems when a later scanf is executed as it will take what is still in the buffer
How do we printout only some of the string?
Left and Right Justify
Run this program and see what you understand.
#include <stdio.h>
/* 08-05.c display only some of the string */
voidpause (void){ /* this function allows me to pause on the screen */
charwaits;
while (getchar() != '\n');
printf ("\n\n\nPause .... hit any key\n\n\n");
scanf("%c", &waits);
while (getchar() != '\n');
}
int main () {
charname1 [11]="1234567890"; // load 10 characters into name1
printf("\n\n\n");
printf("\nDisplay value in name1 [ %s ]", name1);
printf("\nDisplay 3 values using 3s [ %3s ]", name1);
printf("\n\n NOTICE that 3s did not control the printout, the null does");
printf("\n\nDisplay 3 values using .3s [ %.3s ]", name1);
pause();
printf("\n\nDisplay 3 values using 10.3s [ %10.3s ]", name1);
printf("\nDisplay 3 values using -10.3s [ %-10.3s ]", name1);
printf("\n\nNotice what worked and how to do left and right justified\n");
pause();
printf("\nDisplay 3 values using 5.3s [ %5.3s ]", name1);
printf("\nDisplay 3 values using -5.3s [ %-5.3s ]", name1);
printf("\n\nNotice left and right justified are a smaller size of field\n\n\n");
return 0;
}
How to output a string
One character at a time
1for (ctr = 0; a_string[ctr] ; ctr++) {
printf(“%c”, a_string[ctr]);
}
NOTE:strange termination statement
a_string[0] or any index value points to a location with content
any content will result in the statement evaluated as TRUE
except the null character stored at the end of the string evaluates to FALSE
How to output the whole string
2printf(a_string); The first argument to printf is a string
How to do it the way we have in the past
3printf(“\n %s” , a_string);
String Functions you need to know
strcpy()
strcat()
All require the string.h
strcmp()
strlen()
strcpy()
Copies one string to another, therefore, needs 2 arguments
Caution on the order of the argumentsTO .... FROM
strcpy( to , from );
chara_string[20],
b_string[20];
strcpy(a_string, “hello”);to A -- from a string in quotes
strcpy(b_string , a_string);to B -- from A
NOTE:Again no bounds check, so receiver should be the same or bigger than the from string
/*08-strcpy.c demonstrates strcpy */
#include <stdio.h>
#include <string.h>
voidpause (void){ // this function allows me to pause on the screen
printf ("\n\n\nPause .... hit any key\n\n\n");
while (getchar() != '\n'); // clears the buffer
}
main()
{
char a_string[6]; /* holds 5 maximum plus null */
printf ("\n\nThe array called a_string has no initial value %s ", a_string);
pause();
strcpy(a_string, "hello");
printf("\n\n\nThe strcpy function copied hello to a_string\n\n");
printf(a_string);
printf("\n\n\n");
pause();
}
strcat()
It adds the contents of the FROM to the contents of the TO
CAUTION: The TO must be large enough to hold its current contents and the appending or concatenating contents of FROM
charx_string [80];
strcpy (x_string, “First I learn by reading”);
strcat (x_string, “ and then by trying ALL the programs”);
printf (x_string);
/*08-strcat.c*/
#include <stdio.h>
#include <string.h> // don't forget this header
main()
{
char x_string[80];
strcpy(x_string, "First I learn by reading");
strcat(x_string, " and then by trying ALL the programs");
printf("\n\n\n");
printf(x_string);
printf("\n\n\n");
}
08-00-Strings of characters – rt 14 March 20121
strcmp()
Compares 2 strings
strcmp(s1, s2);
If the strings are the samereturns 0
If s1 is greater than s2returns greater than 0
If s1 is less than s2returns less than zero
The comparison is NOT based on length
The comparison is done lexicographically -- dictionary order
a is less than b in the alphabet - see ASCII codes
therefore a is lowest and z is highest
example: all is less than none
printf(“%d”, strcmp(“one”, “one”) ); -- what is printed
NOTE: If you were compare two letters a and b the difference is 1.
The difference between a and c is a difference of 2 etc.
Try this program.
#include <stdio.h>
#include <string.h>
/* 08-strcmp.c display only some of the string */
main () {
charname1 [5]="aaaa", name2[5]="aaaa", name3[5]="xxxx";
int x;
printf("\n\n\n");
printf("\nDisplay value in name1 [ %s ]", name1);
printf("\nDisplay value in name2 [ %s ]", name2);
printf("\nDisplay value in name3 [ %s ]", name3);
printf("\n\n\n");
x = strcmp(name1,name2);
printf("\n\nComparing name1 to name2 gets the answer %d", x);
x = strcmp(name1,name3);
printf("\n\nComparing name1 to name3 gets the answer %d\n\n",x);
x = strcmp(name3,name1);
printf("\n\nComparing name3 to name1 gets the answer %d\n\n",x);
}
strlen()
Returns the length of the string NOT including the null terminator
strlen(x_string);
#include <stdio.h>
#include <string.h>
/* strlen.c display the length of the string */
main (){
char name1[5]="aa",
name2[5]="aaa",
name3[5]="xxxx";
int x;
printf ("\n\n\n");
printf("\nDisplay value in name1 [ %s ]", name1);
printf("\nDisplay value in name2 [ %s ]", name2);
printf("\nDisplay value in name3 [ %s ]", name3);
printf("\n\n\n");
x = strlen(name1);
printf("\n\nLength of string name1 %d\n\n", x);
// NOTE: the different way of burying a function in a function
printf("Length of each is %d %d %d \n\n", strlen(name1), strlen(name2), strlen(name3));
}
Arrays (numeric) with functions
Here is a program that has a hard coded array, calls a function to make changes. Cut and paste it and make other changes to get practice
// change data in an array inside a function
// demonstrates passing data back and forth
void add_2_array (int score []);
main() {
intscore [5]={1,2,3,4,5},
ctr;
for ( ctr = 0 ; ctr < 5 ; ctr ++ ) {
printf("\nDisplay scores loaded in main to start ==> the score[%d] %d",ctr, score[ctr]);
}
add_2_array (score);
for ( ctr = 0 ; ctr < 5 ; ctr ++ ) {
printf("\nMain AFTER changed by function==> the score[%d] %d",ctr, score[ctr]);
}
getchar();
getchar();
}/* END OF MAIN */
void add_2_array (int score []) {
int x;
for (x=0;x<5;x++) {
printf ("\n in function BEFORE change %d", score[x]);
}
for (x=0;x<5;x++) {
score[x] = score [x] +12;
printf ("\n in function %d", score[x]);
}
}
Do the same for character arrays
Changed in a function
You can use the following code to start with or you can write your own.
int main() {
intname [5] = {“abcde”},
ctr;
for ( ctr = 0 ; ctr < 5 ; ctr ++ ) {
printf("\nCharacters before function=> the name[%d] %d",ctr, name[ctr]);
}
Change_name (name);
for ( ctr = 0 ; ctr < 5 ; ctr ++ ) {
printf("\nInn main AFTER changed by function==> the name[%d] %d",ctr, name[ctr]);
}
getchar();
getchar();
return 0;
}
08-00-Strings of characters – rt 14 March 20121