Use the 2D pointer without allocation
In the following C code, I allocate, initialize, and define values for **c. I want **assign to take **c's values. But I only declare a 2D pointer for **assign, no memory allocation, and the code runs successfully and print out the same results as **c. I don't know why?
int main(){
int i,j;
float **c=NULL, **assign=NULL;
c = (float **)calloc(2,sizeof(float *));
for (i=0;i<2;i++){
c[i] = (float *)calloc(3,sizeof(float));
}
for (i=0;i<2;i++){
for (j=0;j<3;j++){
c[i][j] = i+j;
printf("c[%d][%d]=%fn",i,j,c[i][j]);
}
}
assign = c;
for (i=0;i<2;i++){
for (j=0;j<3;j++){
printf("assign[%d][%d]=%fn",i,j,assign[i][j]);
}
}
return 0;
}
c pointers dynamic-memory-allocation
edited Jan 19 at 6:26
Mohammadreza Farahani
2,02531424
asked Jan 19 at 3:11
jamesjames
61
add a comment |
In the following C code, I allocate, initialize, and define values for **c. I want **assign to take **c's values. But I only declare a 2D pointer for **assign, no memory allocation, and the code runs successfully and print out the same results as **c. I don't know why?
int main(){
int i,j;
float **c=NULL, **assign=NULL;
c = (float **)calloc(2,sizeof(float *));
for (i=0;i<2;i++){
c[i] = (float *)calloc(3,sizeof(float));
}
for (i=0;i<2;i++){
for (j=0;j<3;j++){
c[i][j] = i+j;
printf("c[%d][%d]=%fn",i,j,c[i][j]);
}
}
assign = c;
for (i=0;i<2;i++){
for (j=0;j<3;j++){
printf("assign[%d][%d]=%fn",i,j,assign[i][j]);
}
}
return 0;
}
c pointers dynamic-memory-allocation
edited Jan 19 at 6:26
Mohammadreza Farahani
2,02531424
asked Jan 19 at 3:11
jamesjames
61
candassignare pointers, not arrays. Code is simply assigning a pointer withassign = c;.
– chux
Jan 19 at 3:18
1
Two pointers can point to the same memory. As long as the memory being pointed to has been properly allocated this is perfectly safe. You don't have to allocate separately for each pointer. Having said that, note that in C saying that "the code runs successfully" is not always proof that the code is correct. Undefined behavior often looks like code that "runs successfully" most of the time.
– John Coleman
Jan 19 at 3:20
@JohnColeman So you mean the code is correct, but my words are not reasonable?
– james
Jan 19 at 3:51
@james I don't know if your code is correct. I didn't run it or think about it very deeply. Nothing jumps out at me. I just wanted to sound a note of caution to not read too much into the fact that the code seems to be working.
– John Coleman
Jan 19 at 4:17
add a comment |
In the following C code, I allocate, initialize, and define values for **c. I want **assign to take **c's values. But I only declare a 2D pointer for **assign, no memory allocation, and the code runs successfully and print out the same results as **c. I don't know why?
int main(){
int i,j;
float **c=NULL, **assign=NULL;
c = (float **)calloc(2,sizeof(float *));
for (i=0;i<2;i++){
c[i] = (float *)calloc(3,sizeof(float));
}
for (i=0;i<2;i++){
for (j=0;j<3;j++){
c[i][j] = i+j;
printf("c[%d][%d]=%fn",i,j,c[i][j]);
}
}
assign = c;
for (i=0;i<2;i++){
for (j=0;j<3;j++){
printf("assign[%d][%d]=%fn",i,j,assign[i][j]);
}
}
return 0;
}
c pointers dynamic-memory-allocation
edited Jan 19 at 6:26
Mohammadreza Farahani
2,02531424
asked Jan 19 at 3:11
jamesjames
61
In the following C code, I allocate, initialize, and define values for **c. I want **assign to take **c's values. But I only declare a 2D pointer for **assign, no memory allocation, and the code runs successfully and print out the same results as **c. I don't know why?
int main(){
int i,j;
float **c=NULL, **assign=NULL;
c = (float **)calloc(2,sizeof(float *));
for (i=0;i<2;i++){
c[i] = (float *)calloc(3,sizeof(float));
}
for (i=0;i<2;i++){
for (j=0;j<3;j++){
c[i][j] = i+j;
printf("c[%d][%d]=%fn",i,j,c[i][j]);
}
}
assign = c;
for (i=0;i<2;i++){
for (j=0;j<3;j++){
printf("assign[%d][%d]=%fn",i,j,assign[i][j]);
}
}
return 0;
}
c pointers dynamic-memory-allocation
c pointers dynamic-memory-allocation
edited Jan 19 at 6:26
Mohammadreza Farahani
2,02531424
asked Jan 19 at 3:11
jamesjames
61
edited Jan 19 at 6:26
Mohammadreza Farahani
2,02531424
asked Jan 19 at 3:11
jamesjames
61
edited Jan 19 at 6:26
Mohammadreza Farahani
2,02531424
edited Jan 19 at 6:26
Mohammadreza Farahani
2,02531424
edited Jan 19 at 6:26
Mohammadreza Farahani
2,02531424
2,02531424
asked Jan 19 at 3:11
jamesjames
61
asked Jan 19 at 3:11
jamesjames
61
asked Jan 19 at 3:11
jamesjames
61
61
candassignare pointers, not arrays. Code is simply assigning a pointer withassign = c;.
– chux
Jan 19 at 3:18
1
Two pointers can point to the same memory. As long as the memory being pointed to has been properly allocated this is perfectly safe. You don't have to allocate separately for each pointer. Having said that, note that in C saying that "the code runs successfully" is not always proof that the code is correct. Undefined behavior often looks like code that "runs successfully" most of the time.
– John Coleman
Jan 19 at 3:20
@JohnColeman So you mean the code is correct, but my words are not reasonable?
– james
Jan 19 at 3:51
@james I don't know if your code is correct. I didn't run it or think about it very deeply. Nothing jumps out at me. I just wanted to sound a note of caution to not read too much into the fact that the code seems to be working.
– John Coleman
Jan 19 at 4:17
add a comment |
candassignare pointers, not arrays. Code is simply assigning a pointer withassign = c;.
– chux
Jan 19 at 3:18
1
Two pointers can point to the same memory. As long as the memory being pointed to has been properly allocated this is perfectly safe. You don't have to allocate separately for each pointer. Having said that, note that in C saying that "the code runs successfully" is not always proof that the code is correct. Undefined behavior often looks like code that "runs successfully" most of the time.
– John Coleman
Jan 19 at 3:20
@JohnColeman So you mean the code is correct, but my words are not reasonable?
– james
Jan 19 at 3:51
@james I don't know if your code is correct. I didn't run it or think about it very deeply. Nothing jumps out at me. I just wanted to sound a note of caution to not read too much into the fact that the code seems to be working.
– John Coleman
Jan 19 at 4:17
c and assign are pointers, not arrays. Code is simply assigning a pointer with assign = c;.– chux
Jan 19 at 3:18
c and assign are pointers, not arrays. Code is simply assigning a pointer with assign = c;.– chux
Jan 19 at 3:18
1
1
Two pointers can point to the same memory. As long as the memory being pointed to has been properly allocated this is perfectly safe. You don't have to allocate separately for each pointer. Having said that, note that in C saying that "the code runs successfully" is not always proof that the code is correct. Undefined behavior often looks like code that "runs successfully" most of the time.
– John Coleman
Jan 19 at 3:20
Two pointers can point to the same memory. As long as the memory being pointed to has been properly allocated this is perfectly safe. You don't have to allocate separately for each pointer. Having said that, note that in C saying that "the code runs successfully" is not always proof that the code is correct. Undefined behavior often looks like code that "runs successfully" most of the time.
– John Coleman
Jan 19 at 3:20
@JohnColeman So you mean the code is correct, but my words are not reasonable?
– james
Jan 19 at 3:51
@JohnColeman So you mean the code is correct, but my words are not reasonable?
– james
Jan 19 at 3:51
@james I don't know if your code is correct. I didn't run it or think about it very deeply. Nothing jumps out at me. I just wanted to sound a note of caution to not read too much into the fact that the code seems to be working.
– John Coleman
Jan 19 at 4:17
@james I don't know if your code is correct. I didn't run it or think about it very deeply. Nothing jumps out at me. I just wanted to sound a note of caution to not read too much into the fact that the code seems to be working.
– John Coleman
Jan 19 at 4:17
add a comment |
1 Answer
1
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oldest
votes
A pointer is simply a normal variable that holds the address of something else as its value. In other words, a pointer points to the address where something else can be found.
When you allocate:
c = calloc(2,sizeof(float *));
You are assigning the starting address for the new block of memory to c. In other words c points to the location in memory where the first (of two) pointers you allocated can be found.
When you assign (verb):
assign = c;
You are setting the value held by assign to the value held by c. (and what does c hold? -- the address of the block of memory you allocated with calloc). So assign now holds the same address as c, e.g. assign now points to the first (of two) pointers you allocated. So assign and c now both hold the same address as their value and you can use either one to reference what is stored there.
note: there is no need to cast the return of malloc, it is unnecessary. See: Do I cast the result of malloc?. Further, if you use the derefernced pointer to set the type-size for the allocation, you eliminate the chance of getting it wrong, e.g.
c = calloc (2, sizeof *c);
answered Jan 19 at 4:51
David C. RankinDavid C. Rankin
41.3k32748
add a comment |
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A pointer is simply a normal variable that holds the address of something else as its value. In other words, a pointer points to the address where something else can be found.
When you allocate:
c = calloc(2,sizeof(float *));
You are assigning the starting address for the new block of memory to c. In other words c points to the location in memory where the first (of two) pointers you allocated can be found.
When you assign (verb):
assign = c;
You are setting the value held by assign to the value held by c. (and what does c hold? -- the address of the block of memory you allocated with calloc). So assign now holds the same address as c, e.g. assign now points to the first (of two) pointers you allocated. So assign and c now both hold the same address as their value and you can use either one to reference what is stored there.
note: there is no need to cast the return of malloc, it is unnecessary. See: Do I cast the result of malloc?. Further, if you use the derefernced pointer to set the type-size for the allocation, you eliminate the chance of getting it wrong, e.g.
c = calloc (2, sizeof *c);
answered Jan 19 at 4:51
David C. RankinDavid C. Rankin
41.3k32748
add a comment |
A pointer is simply a normal variable that holds the address of something else as its value. In other words, a pointer points to the address where something else can be found.
When you allocate:
c = calloc(2,sizeof(float *));
You are assigning the starting address for the new block of memory to c. In other words c points to the location in memory where the first (of two) pointers you allocated can be found.
When you assign (verb):
assign = c;
You are setting the value held by assign to the value held by c. (and what does c hold? -- the address of the block of memory you allocated with calloc). So assign now holds the same address as c, e.g. assign now points to the first (of two) pointers you allocated. So assign and c now both hold the same address as their value and you can use either one to reference what is stored there.
note: there is no need to cast the return of malloc, it is unnecessary. See: Do I cast the result of malloc?. Further, if you use the derefernced pointer to set the type-size for the allocation, you eliminate the chance of getting it wrong, e.g.
c = calloc (2, sizeof *c);
answered Jan 19 at 4:51
David C. RankinDavid C. Rankin
41.3k32748
add a comment |
A pointer is simply a normal variable that holds the address of something else as its value. In other words, a pointer points to the address where something else can be found.
When you allocate:
c = calloc(2,sizeof(float *));
You are assigning the starting address for the new block of memory to c. In other words c points to the location in memory where the first (of two) pointers you allocated can be found.
When you assign (verb):
assign = c;
You are setting the value held by assign to the value held by c. (and what does c hold? -- the address of the block of memory you allocated with calloc). So assign now holds the same address as c, e.g. assign now points to the first (of two) pointers you allocated. So assign and c now both hold the same address as their value and you can use either one to reference what is stored there.
note: there is no need to cast the return of malloc, it is unnecessary. See: Do I cast the result of malloc?. Further, if you use the derefernced pointer to set the type-size for the allocation, you eliminate the chance of getting it wrong, e.g.
c = calloc (2, sizeof *c);
answered Jan 19 at 4:51
David C. RankinDavid C. Rankin
41.3k32748
A pointer is simply a normal variable that holds the address of something else as its value. In other words, a pointer points to the address where something else can be found.
When you allocate:
c = calloc(2,sizeof(float *));
You are assigning the starting address for the new block of memory to c. In other words c points to the location in memory where the first (of two) pointers you allocated can be found.
When you assign (verb):
assign = c;
You are setting the value held by assign to the value held by c. (and what does c hold? -- the address of the block of memory you allocated with calloc). So assign now holds the same address as c, e.g. assign now points to the first (of two) pointers you allocated. So assign and c now both hold the same address as their value and you can use either one to reference what is stored there.
note: there is no need to cast the return of malloc, it is unnecessary. See: Do I cast the result of malloc?. Further, if you use the derefernced pointer to set the type-size for the allocation, you eliminate the chance of getting it wrong, e.g.
c = calloc (2, sizeof *c);
answered Jan 19 at 4:51
David C. RankinDavid C. Rankin
41.3k32748
answered Jan 19 at 4:51
David C. RankinDavid C. Rankin
41.3k32748
answered Jan 19 at 4:51
David C. RankinDavid C. Rankin
41.3k32748
answered Jan 19 at 4:51
David C. RankinDavid C. Rankin
41.3k32748
41.3k32748
add a comment |
add a comment |
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candassignare pointers, not arrays. Code is simply assigning a pointer withassign = c;.– chux
Jan 19 at 3:18
1
Two pointers can point to the same memory. As long as the memory being pointed to has been properly allocated this is perfectly safe. You don't have to allocate separately for each pointer. Having said that, note that in C saying that "the code runs successfully" is not always proof that the code is correct. Undefined behavior often looks like code that "runs successfully" most of the time.
– John Coleman
Jan 19 at 3:20
@JohnColeman So you mean the code is correct, but my words are not reasonable?
– james
Jan 19 at 3:51
@james I don't know if your code is correct. I didn't run it or think about it very deeply. Nothing jumps out at me. I just wanted to sound a note of caution to not read too much into the fact that the code seems to be working.
– John Coleman
Jan 19 at 4:17