arrays

Question

int main()
{
    int a[30];
    printf("enter string:");
    scanf("%s",a);
    printf("%c\n",a[2]);

}

take a string "123456" tell o/p

Answer 1

In this program u r entering 123456 as input.

But scanf is not storing the input value

So, output of printf will be empty

Comments

instead of a[2] give a[0] now tel what is answer?

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int a[30];

If size of int=4byte and char = 1 byte.

Address allocated for variable a = 30*4= 120 byte.

scanf("%s",a);

Input string = "123456";

If starting address of a = 100; 

then value at address 100 = 1

 value at address 101 = 2

 value at address 102 = 3

 value at address 103 = 4

 value at address 104 = 5

 value at address 105 = 6. 

 value at address 106 = '\0'

---

printf("%c\n",a[2]);

Since data type stored in array is of integer type, but value printed should be only character  

a[0] = Value at address [ 100, 101, 102, 103] but, printf("%c\n",a[0]) will print only value at address 100.;     

a[1] = Value at address [ 104, 105, 106, 107],

but printf("%c\n",a[1]) will print only value at address 104.;     

a[2] = Value at address [ 108, 109, 110, 111].

but printf("%c\n",a[2]) will print only value at address 108.;     

So ans should be garbage value or zero (if default value stored in array is zero).

        

---

nice explaination....

Answer 2

please make me correct , if i am wrong

 

3 rd statement ,  scanf,

compilation problem because format specifier of string (%s) except only char values , but here no char pointer is defined ,

Comments

No, format specifier won't cause a compilation error. Because this is inside a string and requires processing the string which happens at runtime. Some compilers just warns if it finds that the formats are not matching.