![\"\"](\"https:\/\/c-for-dummies.com\/blog\/wp-content\/uploads\/2024\/10\/1026-figure1.png\")
Figure 1. Division and its jargon.<\/p><\/div>\n
With division, you have two values: The numerator or dividend, and the denominator or divisor. The numerator is on top. The denominator, like all demons, goes below. (That’s how I remember it.) The “gazinta” format for division is also shown in Figure 1 (item C). This is the form taught first in school, which is why the remainder becomes a thing.<\/p>\n
In programming, the If you desire to discover the remainder, you can pull a few tricks. The first one is from a blog post<\/a> on the div()<\/em> function, which returns a structure containing the quotient and remainder of a division operation. When you just need to know the remainder, however, you can use the aptly-named remainder()<\/em> function. Here’s its man<\/em> page definition:<\/p>\n This function requires including the The following code demonstrates the remainder()<\/em> function.<\/p>\n The function prompts for two double<\/em> values, which are then thrust into the remainder()<\/em> function:<\/p>\n A printf()<\/em> statement outputs the results, which include standard division to obtain the precise value. But the value of variable This test run threw me:<\/p>\n The result of 29 ÷ 3 is 9 with 2 remaining. But the -1.00 result is accurate; it’s the value 3 minus 1, which is 2. In fact, the documentation for the remainder()<\/em> function explains the result like this:<\/p>\n Value The -1.00 result makes sense given the parameters of the remainder()<\/em> function.<\/p>\n I still prefer the div()<\/em> function to obtain a more realistic answer. But the remainder()<\/em> function remains available, plus it works with real numbers and not just integers.<\/p>\n","protected":false},"excerpt":{"rendered":" Not every value divides evenly into another. Continue reading \/<\/code> operator handles division. In my C programming books, I recommend using float<\/em> or double<\/em> values to obtain the accurate results. When you use integers, the result is truncated.<\/p>\ndouble remainder(double x, double y)<\/code><\/a><\/p>\nmath.h<\/code> header file. In Linux, you must specify the -lm<\/code> switch when compiling to link in the math library.<\/p>\n2024_10_26-Lesson.c<\/a><\/h3>\n
\r\n#include <stdio.h>\r\n#include <math.h>\r\n\r\nint main()\r\n{\r\n double dividend,divisor,r;\r\n\r\n printf(\"Enter the dividend (numerator): \");\r\n scanf(\"%lf\",÷nd);\r\n printf(\"Enter the divisor (denominator): \");\r\n scanf(\"%lf\",&divisor);\r\n\r\n r = remainder(dividend,divisor);\r\n printf(\"%.2f\/%.2f = %.2f with %.2f remainder\\n\",\r\n dividend,\r\n divisor,\r\n dividend\/divisor,\r\n r\r\n );\r\n\r\n return 0;\r\n}<\/pre>\nr = remainder(dividend,divisor);<\/code><\/p>\nr<\/code> is what’s left over — the remainder. Here’s a sample run:<\/p>\nEnter the dividend (numerator): 87.0
\nEnter the divisor (denominator): 6.0
\n87.00\/6.00 = 14.50 with 3.00 remainder<\/code><\/p>\nEnter the dividend (numerator): 29.0
\nEnter the divisor (denominator): 3.0
\n29.00\/3.00 = 9.67 with -1.00 remainder<\/code><\/p>\nx-n*y<\/code><\/p>\nx<\/code> is the dividend and y<\/code> is the divisor, with n<\/code> as the result of x\/y<\/code> rounded to the nearest integer. So:<\/p>\n29-(29\/3)*3 = 29-(10)*3 = 29-30 = -1<\/code><\/p>\n