关于输出的说明

Question

Can anyone please give me the explanation of output

#include<stdio.h>
int main(void){
    int i;
    char *a[]={ "This is first line",
                "This is second line",
                "This is third line",
                "This is fourth line",
                "This is fifth line",
                "This is sixth line",
                "This is seventh line end"};
    printf("%d\n",sizeof(a));
    printf("%d\n",sizeof(*a[0]));
    for(i=0;i<=sizeof(a[0]);i++){
        printf("%s\n",a[i]);
    }
}

Output:

28
1
This is first line
This is second line
This is third line
This is fourth line
This is fifth line

Answer 1

Type of *a[0] is char and type of a[0] is char* (pointer).

sizeof(char) == 1 , sizeof(char*) == 4 .

Type of a is char*[] , sizeof(a) == (7 * sizeof(char*)) == 28 .

Answer 2

Output:

28

a is an array of 7 pointers; each pointer has size 4 on your system. Thus the 28.

1

*a is the same as a[0] , so a[0][0] , *a[0] , **a and (*a)[0] are all equivalent: it is the first character of the first string.

After that, you should get the seven lines.

This is first line
This is second line
This is third line
This is fourth line
This is fifth line

Wait, what? Mmm...

for(i=0;i<=sizeof(a[0]);i++){

seems utterly wrong to me:

for(i=0;i<sizeof(a)/sizeof(a[0]);i++){

should be better. Why? i<=sizeof(a[0]) is i<=4 , giving 5 lines.

However, i<sizeof(a)/sizeof(a[0]); is i < 28/4 , which is 7, the number of elements in the array.

Answer 3

28 - try to print out sizeof(char*), it will be 4 (bytes) so if you have 7 string literals youll have 4*7 = 28 bytes occupied with your array of char* pointers

1 - *a[0] is a size of a single character at address specified by first pointer in your array.

But what is weird that you're using sizeof(a[0]) to get number of your strings. sizeof(a[0]) is equal to sizeof(char*) which is again 4 (bytes). so your for cycle will print out 5 strings because it's equal to this:

for(i=0;i<=4;i++){

It will run 0,1,2,3, 4 included. To print out all strings you should use:

for(i=0;i<sizeof(a)/sizeof(char*);i++){
    printf("%s\n",a[i]);
}

Answer 4

char* a[] is an array of pointers to strings

sizeof will give the number of bytes that the array has but will not tell how many character pointers, to get that you need to divide the total size with the size of one pointer:

sizeof( a ) / sizeof( char * )

an alternative way is to add a NULL pointer in your array

char *a[]={ "This is first line",
            "This is second line",
            "This is third line",
            "This is fourth line",
            "This is fifth line",
            "This is sixth line",
            "This is seventh line end",
            NULL };
printf("%d\n",sizeof(a));
printf("%d\n",sizeof(*a[0]));
for(i=0; a[i] != NULL;i++){
    printf("%s\n",a[i]);
}

Answer 5

This is an array of pointers. In a 32 bit machine, a pointer is 32 bit in size (aka 4 byte). In the first printf() you print the sum of the sizes of all the pointers inside the array. You have 7 lines -> 7*32 = 224 bit = 28 byte. In the second printf() you print the size of the first character of the first line. A char type is 8bit (aka 1 byte in size).

Answer 6

a is array of 7 character pointers. So sizeof(a) is 7*sizeof(int)=28.
a[0] is a character pointer , so value at a[0] (*a[0]) is character whose size is 1.
a[0] is also character pointer which is of type int whose size is 4. So your loop runs from i=0 to i=4. Hence printing 5 strings.