About the hexadecimal result of a-b, where each is a pointer to an array
1
I'm thinking about how the result of the following snippet is 0xc? Shouldn't it be c0-90=30? I run it on ideone.com.
#include <stdio.h>
int main(void) {
int a[10] = {0};
int b[10] = {0};
printf("sizeof(a) = %dn", (int)sizeof(a));
printf("%p, %pn", (void *)(a+10), (void *)(b+10));
printf("a = %p; b = %pn", (void *)a, (void *)b);
printf("a-b = %p", (void *)(a-b));
return 0;
}
Result:
sizeof(a) = 40
0x7ffcabb73ce8, 0x7ffcabb73cb8
a = 0x7ffcabb73cc0; b = 0x7ffcabb73c90
a-b = 0xc
c hex
asked Nov 21 '18 at 5:50
ptr_NEptr_NE
603324
add a comment |
1
I'm thinking about how the result of the following snippet is 0xc? Shouldn't it be c0-90=30? I run it on ideone.com.
#include <stdio.h>
int main(void) {
int a[10] = {0};
int b[10] = {0};
printf("sizeof(a) = %dn", (int)sizeof(a));
printf("%p, %pn", (void *)(a+10), (void *)(b+10));
printf("a = %p; b = %pn", (void *)a, (void *)b);
printf("a-b = %p", (void *)(a-b));
return 0;
}
Result:
sizeof(a) = 40
0x7ffcabb73ce8, 0x7ffcabb73cb8
a = 0x7ffcabb73cc0; b = 0x7ffcabb73c90
a-b = 0xc
c hex
asked Nov 21 '18 at 5:50
ptr_NEptr_NE
603324
It's clear from that output thata+10is the address of the tenthintaftera[0], which is 40 more thana. Unsurprisingly,(a + 10) - ais 10. Consider what that means forb - a(which is technically undefined behaviour, by the way).
– rici
Nov 21 '18 at 6:03
@rici: I thought that would be-(result)...
– ptr_NE
Nov 21 '18 at 6:11
add a comment |
1
1
1
I'm thinking about how the result of the following snippet is 0xc? Shouldn't it be c0-90=30? I run it on ideone.com.
#include <stdio.h>
int main(void) {
int a[10] = {0};
int b[10] = {0};
printf("sizeof(a) = %dn", (int)sizeof(a));
printf("%p, %pn", (void *)(a+10), (void *)(b+10));
printf("a = %p; b = %pn", (void *)a, (void *)b);
printf("a-b = %p", (void *)(a-b));
return 0;
}
Result:
sizeof(a) = 40
0x7ffcabb73ce8, 0x7ffcabb73cb8
a = 0x7ffcabb73cc0; b = 0x7ffcabb73c90
a-b = 0xc
c hex
asked Nov 21 '18 at 5:50
ptr_NEptr_NE
603324
I'm thinking about how the result of the following snippet is 0xc? Shouldn't it be c0-90=30? I run it on ideone.com.
#include <stdio.h>
int main(void) {
int a[10] = {0};
int b[10] = {0};
printf("sizeof(a) = %dn", (int)sizeof(a));
printf("%p, %pn", (void *)(a+10), (void *)(b+10));
printf("a = %p; b = %pn", (void *)a, (void *)b);
printf("a-b = %p", (void *)(a-b));
return 0;
}
Result:
sizeof(a) = 40
0x7ffcabb73ce8, 0x7ffcabb73cb8
a = 0x7ffcabb73cc0; b = 0x7ffcabb73c90
a-b = 0xc
c hex
c hex
asked Nov 21 '18 at 5:50
ptr_NEptr_NE
603324
asked Nov 21 '18 at 5:50
ptr_NEptr_NE
603324
asked Nov 21 '18 at 5:50
ptr_NEptr_NE
603324
asked Nov 21 '18 at 5:50
ptr_NEptr_NE
603324
asked Nov 21 '18 at 5:50
ptr_NEptr_NE
603324
603324
It's clear from that output thata+10is the address of the tenthintaftera[0], which is 40 more thana. Unsurprisingly,(a + 10) - ais 10. Consider what that means forb - a(which is technically undefined behaviour, by the way).
– rici
Nov 21 '18 at 6:03
@rici: I thought that would be-(result)...
– ptr_NE
Nov 21 '18 at 6:11
add a comment |
It's clear from that output thata+10is the address of the tenthintaftera[0], which is 40 more thana. Unsurprisingly,(a + 10) - ais 10. Consider what that means forb - a(which is technically undefined behaviour, by the way).
– rici
Nov 21 '18 at 6:03
@rici: I thought that would be-(result)...
– ptr_NE
Nov 21 '18 at 6:11
It's clear from that output that
a+10 is the address of the tenth int aftera[0], which is 40 more than a. Unsurprisingly, (a + 10) - a is 10. Consider what that means for b - a(which is technically undefined behaviour, by the way).– rici
Nov 21 '18 at 6:03
It's clear from that output that
a+10 is the address of the tenth int aftera[0], which is 40 more than a. Unsurprisingly, (a + 10) - a is 10. Consider what that means for b - a(which is technically undefined behaviour, by the way).– rici
Nov 21 '18 at 6:03
@rici: I thought that would be
-(result)...– ptr_NE
Nov 21 '18 at 6:11
@rici: I thought that would be
-(result)...– ptr_NE
Nov 21 '18 at 6:11
add a comment |
1 Answer
1
active
oldest
votes
2
When you subtract pointers you get the difference in locations not difference in bytes
So, you get 0x7ffcabb73cc0 - 0x7ffcabb73c90 = 0x30 bytes
Dividing this by 4 bytes per integer, you get 0x0c integers
answered Nov 21 '18 at 6:03
Rishikesh RajeRishikesh Raje
5,4591826
0x0c=12(decimal) so the compiler adds 2-integer distance between the start ofaand the end ofbright?
– ptr_NE
Nov 21 '18 at 6:09
2
@ptr_user7813604 - That is dependent on the compiler and/or memory and is not a requirement of the C language.
– Rishikesh Raje
Nov 21 '18 at 6:12
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
When you subtract pointers you get the difference in locations not difference in bytes
So, you get 0x7ffcabb73cc0 - 0x7ffcabb73c90 = 0x30 bytes
Dividing this by 4 bytes per integer, you get 0x0c integers
answered Nov 21 '18 at 6:03
Rishikesh RajeRishikesh Raje
5,4591826
0x0c=12(decimal) so the compiler adds 2-integer distance between the start ofaand the end ofbright?
– ptr_NE
Nov 21 '18 at 6:09
2
@ptr_user7813604 - That is dependent on the compiler and/or memory and is not a requirement of the C language.
– Rishikesh Raje
Nov 21 '18 at 6:12
add a comment |
2
When you subtract pointers you get the difference in locations not difference in bytes
So, you get 0x7ffcabb73cc0 - 0x7ffcabb73c90 = 0x30 bytes
Dividing this by 4 bytes per integer, you get 0x0c integers
answered Nov 21 '18 at 6:03
Rishikesh RajeRishikesh Raje
5,4591826
0x0c=12(decimal) so the compiler adds 2-integer distance between the start ofaand the end ofbright?
– ptr_NE
Nov 21 '18 at 6:09
2
@ptr_user7813604 - That is dependent on the compiler and/or memory and is not a requirement of the C language.
– Rishikesh Raje
Nov 21 '18 at 6:12
add a comment |
2
2
2
When you subtract pointers you get the difference in locations not difference in bytes
So, you get 0x7ffcabb73cc0 - 0x7ffcabb73c90 = 0x30 bytes
Dividing this by 4 bytes per integer, you get 0x0c integers
answered Nov 21 '18 at 6:03
Rishikesh RajeRishikesh Raje
5,4591826
When you subtract pointers you get the difference in locations not difference in bytes
So, you get 0x7ffcabb73cc0 - 0x7ffcabb73c90 = 0x30 bytes
Dividing this by 4 bytes per integer, you get 0x0c integers
answered Nov 21 '18 at 6:03
Rishikesh RajeRishikesh Raje
5,4591826
answered Nov 21 '18 at 6:03
Rishikesh RajeRishikesh Raje
5,4591826
answered Nov 21 '18 at 6:03
Rishikesh RajeRishikesh Raje
5,4591826
answered Nov 21 '18 at 6:03
Rishikesh RajeRishikesh Raje
5,4591826
5,4591826
0x0c=12(decimal) so the compiler adds 2-integer distance between the start ofaand the end ofbright?
– ptr_NE
Nov 21 '18 at 6:09
2
@ptr_user7813604 - That is dependent on the compiler and/or memory and is not a requirement of the C language.
– Rishikesh Raje
Nov 21 '18 at 6:12
add a comment |
0x0c=12(decimal) so the compiler adds 2-integer distance between the start ofaand the end ofbright?
– ptr_NE
Nov 21 '18 at 6:09
2
@ptr_user7813604 - That is dependent on the compiler and/or memory and is not a requirement of the C language.
– Rishikesh Raje
Nov 21 '18 at 6:12
0x0c = 12(decimal) so the compiler adds 2-integer distance between the start of a and the end of b right?– ptr_NE
Nov 21 '18 at 6:09
0x0c = 12(decimal) so the compiler adds 2-integer distance between the start of a and the end of b right?– ptr_NE
Nov 21 '18 at 6:09
2
2
@ptr_user7813604 - That is dependent on the compiler and/or memory and is not a requirement of the C language.
– Rishikesh Raje
Nov 21 '18 at 6:12
@ptr_user7813604 - That is dependent on the compiler and/or memory and is not a requirement of the C language.
– Rishikesh Raje
Nov 21 '18 at 6:12
add a comment |
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It's clear from that output that
a+10is the address of the tenthintaftera[0], which is 40 more thana. Unsurprisingly,(a + 10) - ais 10. Consider what that means forb - a(which is technically undefined behaviour, by the way).– rici
Nov 21 '18 at 6:03
@rici: I thought that would be
-(result)...– ptr_NE
Nov 21 '18 at 6:11