Different answers from strlen and sizeof for Pointer & Array based init of String

Question

Possible Duplicates:
C: differences between pointer and array
Different sizeof results

Basically, I did this...

 char *str1 = "Sanjeev";
 char str2[] = "Sanjeev";
 printf("%d %d\n",strlen(str1),sizeof(str1));    
 printf("%d %d\n",strlen(str2),sizeof(str2));

and my output was

7 4
7 8

I'm not able to give reasons as to why the sizeof str1 is 4. Please explain the difference between these.

@Eugen Constantin Dinca: It's not a dupe because this question is about the sizeof operator and how it works. — Robert S. Barnes, May 08 '11 at 06:21
If I run the code, I get the same output on each line - `7 8 / 7 8`; it is a 64-bit machine, of course. This would probably have misled you about what is going on! — Jonathan Leffler, May 08 '11 at 06:48
@Robert: this question targets a subset of what's covered in the other question's answers. — Eugen Constantin Dinca, May 08 '11 at 07:38

Robert S. Barnes · Accepted Answer · 2011-05-08T06:25:43.373

10

Because sizeof gives you the size in bytes of the data type. The data type of str1 is char* which 4 bytes. In contrast, strlen gives you the length of the string in chars not including the null terminator, thus 7. The sizeof str2 is char array of 8 bytes, which is the number of chars including the null terminator.

See this entry from the C-FAQ: http://c-faq.com/~scs/cclass/krnotes/sx9d.html

Try this:

 char str2[8];
 strncpy(str2, "Sanjeev", 7);
 char *str1 = str2;
 printf("%d %d\n",strlen(str1),sizeof(str1));    
 printf("%d %d\n",strlen(str2),sizeof(str2));

edited May 08 '11 at 06:25

answered May 08 '11 at 06:16

Robert S. Barnes

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`%d` is wrong format specifier to print result of `sizeof`. `sizeof` returns `size_t` which should be displayed using `%zu` specifier introduced by C99 standard – Destructor Feb 20 '17 at 10:37

score 6 · Answer 2 · answered May 08 '11 at 06:15

6

str1 is a pointer to char and size of a pointer variable on your system is 4.
str2 is a array of char on stack, which stores 8 characters, "s","a","n","j","e","e","v","\0" so its size is 8

Important to note that size of pointer to various data type will be same, because they are pointing to different data types but they occupy only size of pointer on that system.

answered May 08 '11 at 06:15

Alok Save

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It is important to add that the size of a pointer is 4 because the author's machine is 32bit, for 64bit machines a pointer is always of size 8. This is the number of bytes needed to represent a memory address in a 32 or 64bit architecture – brita_ Feb 03 '16 at 13:23

score 2 · Answer 3 · answered May 08 '11 at 06:20

sizeof(str1) is sizeof(char*)
whereas sizeof str2 is sizeof(char array)

sizeof(char array) should be equal to strlen(char array) + 1 (NULL termination).
Well if you pass the char array to a function, its size as shown by sizeof will change. The reason is char array is passed as char* only.

score 1 · Answer 4 · answered May 08 '11 at 06:20

 char *str1 = "Sanjeev";
 printf("%d %d\n",strlen(str1),sizeof(str1));

strlen() gives the length of the string which is 7. sizeof() is a compile time operator and sizeof(str1) gives the size of pointer on your implementation (i.e 4).

char str2[] = "Sanjeev";  
printf("%d %d\n",strlen(str2),sizeof(str2));

sizeof(str2) gives the size of the array which is 8*sizeof(char) i.e. 8 [including the NUL terminator]

Different answers from strlen and sizeof for Pointer & Array based init of String

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