Avoiding Common Pitfalls in Learning C Language

1) Assume the beginning of a program is as follows.

#define BOOK “War and Peace”

int main(void){

float coast =12.99;

float percent = 80.0;

Construct a printf() statement using BOOK, coast, and percent to print the following content.

This copy of “War and Peace” sells for $12.99.

That is 80% of list.

Analysis and Answer:

The program requires printing a string with double quotes, so the printf() function needs to include the escape sequence “. Moreover, 80% needs to use the floating-point data percent, and the printout requires adding a modifier, while the percent symbol must be printed using the conversion specifier %%. The specific code is as follows.

*/

#include

#define BOOK “War and Peace”

int main(void){

float coast =12.99;

}

float percent = 80.0;

printf(“This copy of \”%s\” sells for $%.2f.\n”,BOOK,coast);

printf(“That is %.0f%% of list.\n”,percent);

return 0;

2) What conversion specifier should be used to print the following items?

a. A decimal integer with field width equal to its digit count;

b. A hexadecimal integer like 8A, with a field width of 4;

c. A floating-point number like 232.346, with a field width of 10;

d. A floating-point number like 2.33e+002, with a field width of 12;

e. The first 8 characters of a string with a field width of 30.

Analysis and Answer:

a. The conversion specifier for integer data is %d; since the field width is equal to its digit count, no special modifier is needed, so the conversion specifier is %d;

b. The conversion specifier for hexadecimal data (output uppercase characters should use uppercase X) is %X; since the width is 4, using 4 as a modifier, the conversion specifier is %4X;

c. The conversion specifier for floating-point data is %f; for a field width of 10 with 3 decimal places, it should be expressed as 10.3, so the conversion specifier is %10.3f;

d. The conversion specifier for displaying exponent is %e; field width is 12 with 2 decimal places, it should be expressed as 12.2, so the conversion specifier is %12.2e;

e. The conversion specifier for string is %s; with a length of 30, left-aligned, using the modifier -30, the conversion specifier is %-30s;

3) What conversion specifier should be used to print the following items?

a. An unsigned long integer with a field width of 15;

b. A hexadecimal integer like 0x8a, with a field width of 4;

c. A floating-point number like 2.33E+02, with a field width of 12, left-aligned;

d. A floating-point number like +232.346, with a field width of 10;

e. The first 8 characters of a string with a field width of 8.

Analysis and Answer:

a. The conversion specifier for unsigned integer is %u; for a long type field width, the l modifier should be added, the field width is 15, so the conversion specifier is %15lu;

b. The conversion specifier for hexadecimal integer data is %x (output lowercase letters using lowercase x), to output 0X use the # modifier; for a length of 4, use 4 as a modifier, so the conversion specifier is %#4x;

c. The conversion specifier for scientific notation is %E (output uppercase letters using uppercase E), left-aligned using the ‘-‘ modifier, field width is 12, displaying two decimal places using 12.2 modifier, so the conversion specifier is %-12.2E;

d. The conversion specifier for floating-point data is %f; display positive sign using ‘+’ modifier, field width is the character count in the number, meaning no specified field width, showing two decimal places using +0.2 modifier, so the conversion specifier is %+0.2f;

e. The conversion specifier for string is %s; field width is 8, showing the first 8 characters using 8.8 modifier, so the conversion specifier is %8.8s.

4) What conversion specifier should be used to print the following items?

a. A decimal number with a field width of 6, at least 4 digits;

b. An octal integer with field width specified in the parameter list;

c. A character with a field width of 2;

d. A floating-point number like +3.13, with a field width equal to the character count in the number;

e. The first 5 characters of a string with a field width of 7.

Analysis and Answer:

a. The conversion specifier for decimal integer is %d; field width is 6, at least 4 digits, using 6.4 modifier, so the conversion specifier is %6.4d;

b. The conversion specifier for octal data is %o; field width specified by the parameter list, using modifier ‘*’, so the conversion specifier is %*o;

c. The conversion specifier for character data is %c; field width is 2, using modifier 2, so the conversion specifier is %2c;

d. The conversion specifier for floating-point data is %f; display positive sign, field width is equal to the character count in the number, meaning no specified field width, displaying two decimal places using modifier +0.2, so the conversion specifier is %+0.2f;

e. The conversion specifier for string is %s; field width is 7, showing the first 5 characters, left-aligned, using modifier -7.5, so the conversion specifier is %-7.5s.