C doubt

Question

Inside a program loop

if we write

*ch1++

it will give "lvalue required" error

but without any loop if we write even

++*ch1++;

it will not give any lvalue error. Why?

Comments

++*ch1++ = ++(*(ch1++)) which is equal to *(ch1++)= (*(ch1++))+ 1

*ch1++ here lvalue is not present 

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https://stackoverflow.com/questions/22603763/how-to-check-that-behavior-is-undefined-in-c

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no..

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what ???

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that is not my query. Read question carefully

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#include <stdio.h>

int main()
{  int c=10, *ch=&c, i=10;
       *ch++;
     while(i--){
        *ch++;
     }
    printf("Hello, World!\n%d", *ch++);

    return 0;
}

This code isnt giving lvalue required error. I executed it on coding ground as well as online gdp. Isn't it what you were asking for?

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yes, that I know, it will not give lvalue error

But put *ch++ inside while loop. then it will give lvalue error.

Why so? that was my question

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well in my case, its not giving any or probably I didnt get you. Did you mean putting it inside the condition part of while?

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yes, in the condition part

means replace while(i--) with while(*ch++) and check

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int main()
{  int c=10, *ch=&c, i=10;
int count=0;
     while(*ch++){
        
    printf("hey\n");
    if(count++==3)break;
     }

    return 0;
}

No error at all. its printing hey 4 times as expected :/

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:O
but it is giving lvalue
#include <stdio.h>

int main(void) {
char ch1[5]="GATE";
char *ptr-ch1;
while(*ch1++)
{
    printf("%c",*ch1);
}
    return 0;
}

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srestha, ch1 is a constant pointer, you cant change its value -_-

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yes , that I know

But why giving lvalue?

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Because its treating ch1 as a constant, and you need to have a variable on the left hand side so that a value can be assigned to it. So if you write a constant on LHS you get an Lvalue required error.