Variables and Arithmetic Expression
Notes from The C Programming Language
A decimal point in a constant indicates that it is floating point, however, so $5.0/9.0$ i not truncated because it is the ratio of two floating-point values.
printf specification
- %3.0f says that a floating-point number is to be printed at least three characters wide, with no decimal point and no fraction dgits.
- %6.1f describes another number that is to be printed at least six characters wide, with 1 digit after the decimal point.
Width and precision may be omitted from a specification: %6f says that the number is to be at least six characters wide; %.2f specifies two characters after the decimal point, but the width is not constrained.
- %o for octal
- %x for hexadecimal
- %c for character
- %s for character string
- %% for % itself
for statement:
#include <stdio.h> #define LOWER 0 /* lower limit of table */
#define UPPER 300 /* upper limit */
#define STEP 20 /* step size */ /* print Fahrenheit-Celsius table */
main()
{
int fahr; for(fahr = LOWER; fahr <= UPPER; fahr = fahr + STEP)
printf("%3d %6.1f\n", fahr, (5.0/9.0)*(fahr-32));
}
Character input and output
The standard library provides getchar() and putchar() for reading and writing one character at a time. getchar() reads the next input character from a text stream and returns that as its value:
c = getchar();
The function putchar prints a character each time:
putchar(c);
prints the contents of the integer variable c as a character, usually on the screen.
File copy: a program that copies its input to its output one character at a time:
#include <stdio.h> /* copy input to output; 1st version */
main()
{
int c; c = getchar();
while(c != EOF)
{
putchar(c);
c = getchar();
}
}
EOF is defined in <stdio.h>.
As an expression has a value, which is the left hand side after the assignment, the code can be concise:
#include <stdio.h> /* copy input to output; 2nd version */
main()
{
int c; while((c = getchar()) != EOF)
putchar(c);
}
The next program counts characters:
#include <stdio.h> /* count characters in input; 1st version */
main()
{
long nc; nc = 0;
while(getchar() != EOF)
++nc; printf("%ld\n", nc);
}
long integers are at least 32-bits.
It may be possible to cope with even bigger numbers by using a double(double precision float).
#include <stdio.h> /* count characters in input; 2nd version */
main()
{
double nc; for(nc = 0; getchar() != EOF; ++nc)
;
printf("%.0f\n", nc);
}
printf uses %f for both float and double; %.0f suppresses printing of the decimal point and the fraction part, which is zero.
Counts lines:
#include <stdio.h> /* count lines in input */
main()
{
int c, nl; nl = 0;
while((c = getchar()) != EOF)
{
if(c == '\n')
++n1;
printf("%d\n", nl);
}
}
Word counting with loose definition that a word is any sequence of characters that does not contain blank, tab or newline. This is a bare-bones version of the UNIX program wc:
#include <stdio.h> #define IN 1 /* inside a word */
#define OUT 0 /* outside a word*/ /* count lines, words, and characters in input*/
main()
{
int c, nl, nw, nc, state; state = OUT;
nl = nw = nc = 0;
while((c = getchar()) != EOF)
{
++nc;
if(c == '\n')
++nl;
if(c == ' ' || c == '\n' || c == '\t')
state = OUT;
else if(state == OUT)
{
state = IN;
++nw;
}
}
printf("%d %d %d\n", nl, nw, nc);
}
Every time the program encouters the first character of a word, it counts one more word.
Count the number of occurrences of each digit, of white space character(blank, tab, newline), and of all other characters:
#include <stdio.h> /* count digits, white space, others */
main()
{
int c, i, nwhite, nother;
int ndigit[10]; nwhite = nother = 0;
for(i = 0; i < 10; ++i)
ndigit[i] = 0; while((c = getchar()) != EOF)
{
if(c >= '0' && c <= '9')
++ndigit[c - '0'];
else if(c == ' ' || c == '\n' || c == '\t')
++nwhite;
else
++nother; printf("digits =");
for(i = 0; i < 10; ++i)
printf(" %d", ndigit[i]);
printf(", white space = %d, other = %d\n", nwhite, nother);
}
}
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