Marcin Jałbrzykowski - rozwiązanie zadania z argumentami

README

@@ -0,0 +1 @@
+Marcin Jalbrzykowski

vim-challenge/args.c

@@ -0,0 +1,50 @@
+#include <string.h>
+#include <stdlib.h>
+#include <stdio.h>
+#include <stdarg.h>
+
+struct test {
+	int i1, i2;
+	double d;
+	char i3;
+};
+
+struct test* alloc(const char *fmt, ...) { 
+	struct test *out = malloc(sizeof(struct test*));
+	va_list args;
+	va_start(args, fmt);
+
+	int i1_has_value = 0;
+	int i2_has_value = 0;
+
+	for(int i = 0; i < strlen(fmt); i++) {
+		switch(fmt[i]) {
+			case 'i':
+				if(!i1_has_value) {
+					out->i1 = va_arg(args,int);
+					i1_has_value = 1;
+				}
+				else if(!i2_has_value) {
+					out->i2 = va_arg(args,int);
+					i2_has_value = 1;
+				}
+				else 
+					out->i3 = va_arg(args,int);
+				break;
+			case 'd':
+				out->d = va_arg(args,double);
+				break;
+		}
+	}
+
+	va_end(args);
+
+	return out;
+}
+
+int main() {
+	struct test *p = alloc("iidi", 33, 44, 3.14, 55);
+	printf("%d %d %f %d\n", p->i1, p->i2, p->d, p->i3);
+
+	return 0;
+}

vim-challenge/integral.c

@@ -1,34 +1,39 @@
 #include <stdio.h>
+#include <math.h>
 
-float integral(float a, float b, float (*f) (float), int n) {
- int i;
-	float dx = (b - a) / n;
-    float value_of_integral = 0.0;
-
-for (i=0; i<n; i++) {
-float si_a = a + (i)*dx;
-float si_b = a + (i+1)*dx;
-value_of_integral += simpson(si_a, si_b, f, dx);
+double simpson(double a, double b, double (*f) (double), double dx) {
+	double value_of_interval_integral =
+					0.5 * (f(a) + f(b)) / dx;
+	return value_of_interval_integral;
 }
 
-return value_of_integral;
+double integral(double a, double b, double (*f) (double), int n) {
+	int i;
+	double dx = (b - a) / n;
+	double value_of_integral = 0.0;
+
+	for (i=0; i<n; i++) {
+		double subinterval_a = a + (i)*dx;
+		double subinterval_b = a + (i+1)*dx;
+		value_of_integral += subintervalmpson(si_a, si_b, f, dx);
+	}
+
+	return value_of_integral;
 }
 
-float square(float x) {
+double square(double x) {
 	return x*x;
 }
 
-float simpson(float a, float b, float (*f) (float), float dx) {
-	float value_of_interval_integral =
-					0.5 * (f(a) + f(b)) / dx;
-	return value_of_interval_integral;
+double cube(double x) {
+	return x*x*x;
 }
 
 int main(int argc, char** argv) {
-	float twopi = 2 * 3.14;
-	float integral_of_f = integral(0.0, 1.0, square, 1e3);
-	/*float integral_of_sin = integral(0.0, twopi, sin, 10);*/
-	printf("Całka z funkjci f (0.0 - 1.0) = %g\n", integral_of_f);
+	double twopi = 2 * 3.14;
+	double integral_of_f = integral(0.0, 1.0, cube, 1e3);
+	/*double integral_of_sin = integral(0.0, twopi, sin, 10);*/
+	printf("Całka z funkcji f (0.0 - 1.0) = %g\n", integral_of_f);
 	/*printf("Całka z sinus (0 - 2pi) = %g\n", integral_of_sin);*/
 	return 0;
 }