Sequencing UB

sequence-points/README

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+The expectation is that without sequence points between read/writes C will evaluate in a left to right fashion. Also, each modification of a value will complete before the next use of that variable.

sequence-points/p1.c

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+#include <stdio.h>
+#include <stdlib.h>
+
+
+/* comments say what the programmer would expect */
+
+const int nums[32] = {1, 2, 3, 4, 5, 6, 7, 8,
+		      9, 10, 11, 12, 13, 14,
+		      15, 16, 17, 18, 19, 20,
+		      21, 22, 23, 24, 25, 26,
+		      27, 28, 29, 30, 31, 32 };
+
+int main(void) {
+
+	int index = 0;
+
+	index = nums[index++]; //index = nums[0]; index++; -> index = 2;
+	index = nums[index++]; //index = nums[2]; index++; -> index = 4;
+	index = nums[index++]; //index = nums[4]; index++; -> index = 6;
+	index = nums[index++]; //index = nums[6]; index++; -> index = 8;
+	index = nums[index++]; //index = nums[8]; index++; -> index = 10;
+	index = nums[index++]; //index = nums[10]; index++; -> index = 12;
+	printf("index is %d\n", index);
+
+	return EXIT_SUCCESS;
+}

sequence-points/p1.output

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+index is 12

sequence-points/p2.c

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+#include <stdio.h>
+#include <stdlib.h>
+
+int main(void)
+{
+
+	unsigned int x = 15;
+	unsigned int y = 25;
+	unsigned result = 0;
+
+	/* programmer assumes left to right evaluation,
+	   even with unary ops mixed in */
+	/* 16 + 15 + (50 * 50) + 50 + 15 */
+	result = x = y = x++ + x-- + (y*=2 * y) + y + x ;
+
+	printf("result %d\n", result );
+
+	return EXIT_SUCCESS;
+}

sequence-points/p2.output

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+result 2596