Reputation: 21891
What is the difference between memcpy()
and strcpy()
? I tried to find it with the help of a program but both are giving the same output.
int main()
{
char s[5] = { 's', 'a', '\0', 'c', 'h' };
char p[5];
char t[5];
strcpy(p, s);
memcpy(t, s, 5);
printf("sachin p is [%s], t is [%s]", p, t);
return 0;
}
Output
sachin p is [sa], t is [sa]
Upvotes: 99
Views: 138129
Reputation: 11
printf("%s",...)
stops printing the data when null is encountered, so both the outputs are same.
The following code differentiates between strcpy
and memcpy
:
#include<stdio.h>
#include<string.h>
int main()
{
char s[5]={'s','a','\0','c','h'};
char p[5];
char t[5];
int i;
strcpy(p,s);
memcpy(t,s,5);
for(i=0;i<5;i++)
printf("%c",p[i]);
printf("\n");
for(i=0;i<5;i++)
printf("%c",t[i]);
return 0;
}
Upvotes: 1
Reputation: 33187
strcpy
stops when it encounters a NUL ('\0'
) character, memcpy
does not. You do not see the effect here, as %s
in printf also stops at NUL.
Upvotes: 118
Reputation: 73081
The main difference is that memcpy()
always copies the exact number of bytes you specify; strcpy()
, on the other hand, will copy until it reads a NUL (aka 0) byte, and then stop after that.
Upvotes: 6
Reputation: 7161
Because of the null character in your s
string, the printf
won't show anything beyond that. The difference between p
and t
will be in characters 4 and 5. p
won't have any (they'll be garbage) and t
will have the 'c'
and 'h'
.
Upvotes: 4
Reputation: 775
strcpy
stops when it encounters a NULL
or '\0'
memcpy
is usually more efficient than strcpy
, which always scan the data it copiesUpvotes: 3
Reputation: 41
The problem with your test-program is, that the printf()
stops inserting the argument into %s
, when it encounters a null-termination \0
.
So in your output you probably have not noticed, that memcpy()
copied the characters c
and h
as well.
I have seen in GNU glibc-2.24
, that (for x86) strcpy()
just calls memcpy(dest, src, strlen(src) + 1)
.
Upvotes: 3
Reputation: 4880
strcpy
copies character from source to destination one by one until it find NULL or '\0' character in the source.
while((*dst++) = (*src++));
where as memcpy
copies data (not character) from source to destination of given size n, irrespective of data in source.
memcpy
should be used if you know well that source contain other than character. for encrypted data or binary data, memcpy is ideal way to go.
strcpy
is deprecated, so use strncpy
.
Upvotes: 10
Reputation: 25053
what could be done to see this effect
Compile and run this code:
void dump5(char *str);
int main()
{
char s[5]={'s','a','\0','c','h'};
char membuff[5];
char strbuff[5];
memset(membuff, 0, 5); // init both buffers to nulls
memset(strbuff, 0, 5);
strcpy(strbuff,s);
memcpy(membuff,s,5);
dump5(membuff); // show what happened
dump5(strbuff);
return 0;
}
void dump5(char *str)
{
char *p = str;
for (int n = 0; n < 5; ++n)
{
printf("%2.2x ", *p);
++p;
}
printf("\t");
p = str;
for (int n = 0; n < 5; ++n)
{
printf("%c", *p ? *p : ' ');
++p;
}
printf("\n", str);
}
It will produce this output:
73 61 00 63 68 sa ch
73 61 00 00 00 sa
You can see that the "ch" was copied by memcpy()
, but not strcpy()
.
Upvotes: 143
Reputation: 35925
strcpy
terminates when the source string's null terminator is found. memcpy
requires a size parameter be passed. In the case you presented the printf
statement is halting after the null terminator is found for both character arrays, however you will find t[3]
and t[4]
have copied data in them as well.
Upvotes: 14