Simple example below what make array & pointer compatible!! &

Question

Simple example below what make array & pointer compatible!! &

different!!

#include <stdio.h>
#include <string.h>

int main(void)
{
char a[]="hello world";
char *b="hello world";

printf(" a6: %c\n", a[6]);
printf("a6+1: %c\n", a[6+1]);

printf(" b+6: %s\n", b+6);
*b++;*b++; /: for pro!
// b++; b++; // for gcc! & pro!
printf(" b++: %s\n", b);
printf(" b++: %c\n", b[0]); // NOTE
// the above last statement holds incremented pointer!!

printf(" b5: %c\n", b[5]);
printf("b5+1: %c\n", b[5+1]);

printf("la: %lu\n", strlen(a));
printf("lb: %lu\n", strlen(b));

printf(" %x\n", a[11]); // NULL/0: both
printf(" %x\n", b[9]); // 0

// printf(" %c\n", a[12]); // ERROR
// printf(" %c\n", b[25]); // NO-ERROR
// no-error in above line why?

}

Hoping GCC doesn’t throw any exception 😋

0

03.10.2019

4 ответов

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Aakash Saini Автор вопроса

I’ve modified above program to show how increment works!! in pointers further..

0

03.10.2019

swarnim

There aren't any exceptions in C

0

03.10.2019

Chandradeep Dey

and this

0

03.10.2019

Chandradeep Dey · Accepted Answer

line 14 - *b++;*b++; -> useless and insane. b++; b++; would be same but still insane. ++b; ++b; would be more efficient but insane. b += 2; would be sane. line 17 - // the above last statement holds incremented pointer!! - > statements don't hold incremented anything, variables do. line 26 - accessing pointer out of range, invalid line 27 - accessing pointer out of range, using invalid format specification, also invalid line 30 - it is an error. accessing pointer out of range

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Simple example below what make array & pointer compatible!! &

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