Mixing
Xcode debugger showing a property as ?
I'm inspecting an enum which has an init method:
init(cardholder: Cardholder?, topUpSettings: TopUpSettings?, cardUser: VenmoCardUser? = nil) {
switch (cardholder, topUpSettings, cardUser) {
case (nil, _, let cardUser?) where cardUser.isWhitelisted && cardUser.cardType == .paypal:
self = .readyToOnboard
The VenmoCardUser.cardType is itself a VenmoCardType:
public enum VenmoCardType: String, Codable {
case shift = "shift"
case paypal = "paypal"
}
I would expect the cardUser.cardType in this method to be "paypal", but I'm a bit confused by the debugger output:
The debugger inspector (the window on the bottom left) is showing cardUser multiple times; the bottommost occurrence shows the cardUser.cardType as shift, whereas the second-to-last shows it as <invalid>. If I evaluate expression cardUser.cardType in lldb, I seem to get <invalid>.
Can someone perhaps help me understand
- Why I am seeing
cardUsertwice in the debugger - what do the different occurrences represent? - How can the value be
<invalid>? Doesn't theenumhave to be one of the permissible valuesshiftorpaypal?
Update
To answer my first question, I believe that the debugger re-evaluates the variables in scope every time you press the "Step over" button, and outputs them below the ones you saw before. Below, I've added another screenshot in which I restarted the emulator, hit the breakpoint, and pressed "Step over" until the green highlighting moved to the next case.
What still baffles me, however, is that cardUser.cardType starts out as paypal (in the magenta drop-down menu), but after I press "Step over" multiple times, it changes first to <invalid>, and ultimately to shift like in the bottommost occurrence in the original screenshot. How can I tell how or at what point this is happening?
ios swift xcode
edited Jan 3 at 4:15
shim
4,10664780
asked Jan 2 at 21:21
Kurt PeekKurt Peek
10.9k2276162
add a comment |
I'm inspecting an enum which has an init method:
init(cardholder: Cardholder?, topUpSettings: TopUpSettings?, cardUser: VenmoCardUser? = nil) {
switch (cardholder, topUpSettings, cardUser) {
case (nil, _, let cardUser?) where cardUser.isWhitelisted && cardUser.cardType == .paypal:
self = .readyToOnboard
The VenmoCardUser.cardType is itself a VenmoCardType:
public enum VenmoCardType: String, Codable {
case shift = "shift"
case paypal = "paypal"
}
I would expect the cardUser.cardType in this method to be "paypal", but I'm a bit confused by the debugger output:
The debugger inspector (the window on the bottom left) is showing cardUser multiple times; the bottommost occurrence shows the cardUser.cardType as shift, whereas the second-to-last shows it as <invalid>. If I evaluate expression cardUser.cardType in lldb, I seem to get <invalid>.
Can someone perhaps help me understand
- Why I am seeing
cardUsertwice in the debugger - what do the different occurrences represent? - How can the value be
<invalid>? Doesn't theenumhave to be one of the permissible valuesshiftorpaypal?
Update
To answer my first question, I believe that the debugger re-evaluates the variables in scope every time you press the "Step over" button, and outputs them below the ones you saw before. Below, I've added another screenshot in which I restarted the emulator, hit the breakpoint, and pressed "Step over" until the green highlighting moved to the next case.
What still baffles me, however, is that cardUser.cardType starts out as paypal (in the magenta drop-down menu), but after I press "Step over" multiple times, it changes first to <invalid>, and ultimately to shift like in the bottommost occurrence in the original screenshot. How can I tell how or at what point this is happening?
ios swift xcode
edited Jan 3 at 4:15
shim
4,10664780
asked Jan 2 at 21:21
Kurt PeekKurt Peek
10.9k2276162
add a comment |
I'm inspecting an enum which has an init method:
init(cardholder: Cardholder?, topUpSettings: TopUpSettings?, cardUser: VenmoCardUser? = nil) {
switch (cardholder, topUpSettings, cardUser) {
case (nil, _, let cardUser?) where cardUser.isWhitelisted && cardUser.cardType == .paypal:
self = .readyToOnboard
The VenmoCardUser.cardType is itself a VenmoCardType:
public enum VenmoCardType: String, Codable {
case shift = "shift"
case paypal = "paypal"
}
I would expect the cardUser.cardType in this method to be "paypal", but I'm a bit confused by the debugger output:
The debugger inspector (the window on the bottom left) is showing cardUser multiple times; the bottommost occurrence shows the cardUser.cardType as shift, whereas the second-to-last shows it as <invalid>. If I evaluate expression cardUser.cardType in lldb, I seem to get <invalid>.
Can someone perhaps help me understand
- Why I am seeing
cardUsertwice in the debugger - what do the different occurrences represent? - How can the value be
<invalid>? Doesn't theenumhave to be one of the permissible valuesshiftorpaypal?
Update
To answer my first question, I believe that the debugger re-evaluates the variables in scope every time you press the "Step over" button, and outputs them below the ones you saw before. Below, I've added another screenshot in which I restarted the emulator, hit the breakpoint, and pressed "Step over" until the green highlighting moved to the next case.
What still baffles me, however, is that cardUser.cardType starts out as paypal (in the magenta drop-down menu), but after I press "Step over" multiple times, it changes first to <invalid>, and ultimately to shift like in the bottommost occurrence in the original screenshot. How can I tell how or at what point this is happening?
ios swift xcode
edited Jan 3 at 4:15
shim
4,10664780
asked Jan 2 at 21:21
Kurt PeekKurt Peek
10.9k2276162
I'm inspecting an enum which has an init method:
init(cardholder: Cardholder?, topUpSettings: TopUpSettings?, cardUser: VenmoCardUser? = nil) {
switch (cardholder, topUpSettings, cardUser) {
case (nil, _, let cardUser?) where cardUser.isWhitelisted && cardUser.cardType == .paypal:
self = .readyToOnboard
The VenmoCardUser.cardType is itself a VenmoCardType:
public enum VenmoCardType: String, Codable {
case shift = "shift"
case paypal = "paypal"
}
I would expect the cardUser.cardType in this method to be "paypal", but I'm a bit confused by the debugger output:
The debugger inspector (the window on the bottom left) is showing cardUser multiple times; the bottommost occurrence shows the cardUser.cardType as shift, whereas the second-to-last shows it as <invalid>. If I evaluate expression cardUser.cardType in lldb, I seem to get <invalid>.
Can someone perhaps help me understand
- Why I am seeing
cardUsertwice in the debugger - what do the different occurrences represent? - How can the value be
<invalid>? Doesn't theenumhave to be one of the permissible valuesshiftorpaypal?
Update
To answer my first question, I believe that the debugger re-evaluates the variables in scope every time you press the "Step over" button, and outputs them below the ones you saw before. Below, I've added another screenshot in which I restarted the emulator, hit the breakpoint, and pressed "Step over" until the green highlighting moved to the next case.
What still baffles me, however, is that cardUser.cardType starts out as paypal (in the magenta drop-down menu), but after I press "Step over" multiple times, it changes first to <invalid>, and ultimately to shift like in the bottommost occurrence in the original screenshot. How can I tell how or at what point this is happening?
ios swift xcode
ios swift xcode
edited Jan 3 at 4:15
shim
4,10664780
asked Jan 2 at 21:21
Kurt PeekKurt Peek
10.9k2276162
edited Jan 3 at 4:15
shim
4,10664780
asked Jan 2 at 21:21
Kurt PeekKurt Peek
10.9k2276162
edited Jan 3 at 4:15
shim
4,10664780
edited Jan 3 at 4:15
shim
4,10664780
edited Jan 3 at 4:15
shim
4,10664780
4,10664780
asked Jan 2 at 21:21
Kurt PeekKurt Peek
10.9k2276162
asked Jan 2 at 21:21
Kurt PeekKurt Peek
10.9k2276162
asked Jan 2 at 21:21
Kurt PeekKurt Peek
10.9k2276162
10.9k2276162
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
switch (cardholder, topUpSettings, cardUser) {
//BREAK 1
case (nil, _, let cardUser?) where cardUser.isWhitelisted && cardUser.cardType == .paypal:
//BREAK 2 ON THE LINE ABOVE
If we break on 1, we will see that if we print our cardUser it will return some value. In your case, it LOOKS like it should be paypal initially. However, when you step over and break on the next line of case(nil, _, let cardUser?)... you are ON the line of let cardUser? therefore, we don't yet KNOW what it is. You need to go to the NEXT line to determine what cardUser will be. When you step on to the line, it has not yet executed. You are on the line, therefore cardUser has not yet executed, but is about to.
You are seeing 2 cardUser because you are USING 2 cardUser variables
Try changing to this
switch (cardholder, topUpSettings, cardUser) {
case (nil, _, let differentCardUser) ...
You will now see cardUser and differentCardUser. cardUser will have its value and differentCardUser will have its value AFTER you step over the initialization of it.
answered Jan 2 at 22:30
impression7vximpression7vx
54911038
add a comment |
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1 Answer
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oldest
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1 Answer
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active
oldest
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oldest
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oldest
votes
switch (cardholder, topUpSettings, cardUser) {
//BREAK 1
case (nil, _, let cardUser?) where cardUser.isWhitelisted && cardUser.cardType == .paypal:
//BREAK 2 ON THE LINE ABOVE
If we break on 1, we will see that if we print our cardUser it will return some value. In your case, it LOOKS like it should be paypal initially. However, when you step over and break on the next line of case(nil, _, let cardUser?)... you are ON the line of let cardUser? therefore, we don't yet KNOW what it is. You need to go to the NEXT line to determine what cardUser will be. When you step on to the line, it has not yet executed. You are on the line, therefore cardUser has not yet executed, but is about to.
You are seeing 2 cardUser because you are USING 2 cardUser variables
Try changing to this
switch (cardholder, topUpSettings, cardUser) {
case (nil, _, let differentCardUser) ...
You will now see cardUser and differentCardUser. cardUser will have its value and differentCardUser will have its value AFTER you step over the initialization of it.
answered Jan 2 at 22:30
impression7vximpression7vx
54911038
add a comment |
switch (cardholder, topUpSettings, cardUser) {
//BREAK 1
case (nil, _, let cardUser?) where cardUser.isWhitelisted && cardUser.cardType == .paypal:
//BREAK 2 ON THE LINE ABOVE
If we break on 1, we will see that if we print our cardUser it will return some value. In your case, it LOOKS like it should be paypal initially. However, when you step over and break on the next line of case(nil, _, let cardUser?)... you are ON the line of let cardUser? therefore, we don't yet KNOW what it is. You need to go to the NEXT line to determine what cardUser will be. When you step on to the line, it has not yet executed. You are on the line, therefore cardUser has not yet executed, but is about to.
You are seeing 2 cardUser because you are USING 2 cardUser variables
Try changing to this
switch (cardholder, topUpSettings, cardUser) {
case (nil, _, let differentCardUser) ...
You will now see cardUser and differentCardUser. cardUser will have its value and differentCardUser will have its value AFTER you step over the initialization of it.
answered Jan 2 at 22:30
impression7vximpression7vx
54911038
add a comment |
switch (cardholder, topUpSettings, cardUser) {
//BREAK 1
case (nil, _, let cardUser?) where cardUser.isWhitelisted && cardUser.cardType == .paypal:
//BREAK 2 ON THE LINE ABOVE
If we break on 1, we will see that if we print our cardUser it will return some value. In your case, it LOOKS like it should be paypal initially. However, when you step over and break on the next line of case(nil, _, let cardUser?)... you are ON the line of let cardUser? therefore, we don't yet KNOW what it is. You need to go to the NEXT line to determine what cardUser will be. When you step on to the line, it has not yet executed. You are on the line, therefore cardUser has not yet executed, but is about to.
You are seeing 2 cardUser because you are USING 2 cardUser variables
Try changing to this
switch (cardholder, topUpSettings, cardUser) {
case (nil, _, let differentCardUser) ...
You will now see cardUser and differentCardUser. cardUser will have its value and differentCardUser will have its value AFTER you step over the initialization of it.
answered Jan 2 at 22:30
impression7vximpression7vx
54911038
switch (cardholder, topUpSettings, cardUser) {
//BREAK 1
case (nil, _, let cardUser?) where cardUser.isWhitelisted && cardUser.cardType == .paypal:
//BREAK 2 ON THE LINE ABOVE
If we break on 1, we will see that if we print our cardUser it will return some value. In your case, it LOOKS like it should be paypal initially. However, when you step over and break on the next line of case(nil, _, let cardUser?)... you are ON the line of let cardUser? therefore, we don't yet KNOW what it is. You need to go to the NEXT line to determine what cardUser will be. When you step on to the line, it has not yet executed. You are on the line, therefore cardUser has not yet executed, but is about to.
You are seeing 2 cardUser because you are USING 2 cardUser variables
Try changing to this
switch (cardholder, topUpSettings, cardUser) {
case (nil, _, let differentCardUser) ...
You will now see cardUser and differentCardUser. cardUser will have its value and differentCardUser will have its value AFTER you step over the initialization of it.
answered Jan 2 at 22:30
impression7vximpression7vx
54911038
answered Jan 2 at 22:30
impression7vximpression7vx
54911038
answered Jan 2 at 22:30
impression7vximpression7vx
54911038
answered Jan 2 at 22:30
impression7vximpression7vx
54911038
54911038
add a comment |
add a comment |
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