pointer and string part1

[@shahshah-wi9kk]

if("abc"=="abc") printf("Equal strings"); else printf("Unequal strings"); output: Equal strings explanation: - In this case, both "abc" are string literals. - String literals with the same contents might be stored at the same memory location by the compiler, depending on the implementation. - Therefore, the comparison "abc" == "abc" can be true because both literals might point to the same address in memory. - Thus, the output is Equal strings. char x[4] = "abc"; //used size [4] to include null terminator char y[4] = "abc"; if(x==y) printf("equal strings"); else printf("unequal strings"); output: unequal strings explanation: -Here, x and y are character arrays, not string literals. - Each array x and y will have its own separate memory location. - When you use the == operator on arrays, you are comparing their addresses, not their contents. - Since x and y are different arrays, their addresses will not be the same. - Therefore, the comparison x == y is false, and the output is Unequal strings. right approach to compare strings: #include #include int main() { char x[4] = "abc"; // Note: x[4] to include the null terminator '\0' char y[4] = "abc"; // Note: y[4] to include the null terminator '\0' if(strcmp(x, y) == 0) printf("Equal strings"); else printf("Unequal strings"); return 0; }

[@aryushgupta8680]

Thanks buddy!!

[@GateCse-2025]

thanks chatgpt.

[@satyaprakashsingh952]

I have learing your all pointer video sir Too good explained by you

[@sammanluitel3018]

Please make complete c tutorial please with many examples of coding problems using D's algo aswell

[@AmitKhuranaSir]

I will make playlist on C language soon.....

[@atultripathi52]

sir you are wrong at 2.25 "abc"=="abc" returns 1

[@guru_cse]

It is compiler dependent.

[@dherendrasingh4761]

right , me also check on online gdb, getting true condition result

[@dherendrasingh4761]

Sir, i check ABC==ABC concept by program but I found true condition result . How ABC is different from ABC ? #include int main() { char x[5] ="ABCD"; printf("%s ",x); if("ABC" == "ABC") printf("HELLO "); else printf("BYE "); return 0; } O/P ABCD HELLO

[@rndmdpe23]

same happened with me as well, if you do printf("%u %u","string","string"); how many times you want, you ge tthe same address. maybe for same strings the same one created in the memory is used(like in JAVA). but if we would've done like char a[]="ok', b[]="ok". then if(a==b) would result in false. this was all i could understand i hope it helps also correct me if i'm wrong

[@atultripathi52]

also sir, adding [] with abcd while printing gives error message

[@AmitKhuranaSir]

All these are compiler dependent... And i am teaching according to turbo C compiler....